3.1 Solutions to Problems in Special Relativity Theory
1.1
a) A scalar is always the same independent of the inertial frame. The scalar product between the two 4 -vectors vec(A)\vec{A} and vec(B)\vec{B} is zero in the inertial frame of the observer O^(')\mathcal{O}^{\prime} since they point in different coordinate directions. Thus, it is always zero. The two 4 -vectors vec(U)\vec{U} and vec(V)\vec{V} are lightlike in different directions. The scalar product between two lightlike 4 -vectors pointing in different directions is always nonzero. Assume that U^(0)=|alpha|,U^(1)=-|alpha|U^{0}=|\alpha|, U^{1}=-|\alpha| and V^(0)=|beta|,V^(1)=|beta|V^{0}=|\beta|, V^{1}=|\beta|, where alpha,beta!=0\alpha, \beta \neq 0, then we have
In conclusion, the statements 1,2 , and 6 are true.
b) The 4 -vector vec(A)\vec{A} can be proportional to a 4 -velocity vector for a massive particle, since vec(A)\vec{A} is equal to (A^(0),0,0,0)\left(A^{0}, 0,0,0\right) in S^(')S^{\prime} a 4-velocity vector of a massive particle is given by (c,0,0,0)(c, 0,0,0), where cc is the speed of light. The two 4 -vectors vec(U)\vec{U} and vec(V)\vec{V} are proportional to the worldline tangent vector of massless particles, since they are null vectors. Vector vec(B)\vec{B} is spacelike and cannot be the tangent of a worldline.
1.2
a) Let A^(mu)A^{\mu} be the timelike 4 -vector. There exists an inertial coordinate system K^(')K^{\prime} such that A^('mu)=(A^('0),0)A^{\prime \mu}=\left(A^{\prime 0}, 0\right), where A^('0)!=0A^{\prime 0} \neq 0. If the 4 -vector B^(mu)B^{\mu} is orthogonal to A^(mu)A^{\mu}, then A^('mu)B_(mu)^(')=A^('0)B_(0)^(')=0A^{\prime \mu} B_{\mu}^{\prime}=A^{\prime 0} B_{0}^{\prime}=0, which means that B_(0)^(')=B^(0)=0B_{0}^{\prime}=B^{0}=0. Hence, it holds that
which means that B^(mu)B^{\mu} is spacelike. (We assume that B^(mu)≢0B^{\mu} \not \equiv 0.)
b) Let A^(mu)A^{\mu} and B^(mu)B^{\mu} be the 4 -vectors. There exists an inertial coordinate system K^(')K^{\prime} such that A^('mu)=(A^('0),0)A^{\prime \mu}=\left(A^{\prime 0}, \mathbf{0}\right) and B^('mu)=(B^('0),B^('))B^{\prime \mu}=\left(B^{\prime 0}, B^{\prime}\right), where A^('0) > 0A^{\prime 0}>0 and B^(0) > 0B^{0}>0. Therefore, it holds that
because (B^('0))^(2)-B^('2) > 0\left(B^{\prime 0}\right)^{2}-B^{\prime 2}>0, since B^(mu)B^{\mu} is timelike.
c) There exists an inertial coordinate system K^(')K^{\prime} such that the spacelike 4-vector has the components A^('mu)=(0,A^('1),0,0)A^{\prime \mu}=\left(0, A^{\prime 1}, 0,0\right) relative to K^(')K^{\prime}. Thus, we can set
where V_(1)V_{1} and V_(2)V_{2} are the 4 -velocities of the observers. (This may be seen by going to the rest frame of one of the observers, where the 4 -velocity of two other is given by V=gamma_(12)(1,v_("rel "))V=\gamma_{12}\left(1, \boldsymbol{v}_{\text {rel }}\right).) We find that
1.4
a) The correct answer is: Statement 3 (never). This is due to the fact that a photon is traveling at the speed of light relative to all inertial frames. Hence, it is impossible to perform a Lorentz transformation to a frame where it is does not move.
b) The correct answer is: Statement 2 (sometimes). In order to have a rest frame for the center of momentum for a system of two photons, it cannot be moving with
the speed of light. This happens when the two photons are moving in directions parallel to each other. However, in all other configurations, the center of momentum will move at a speed slower than cc, and then, we can find a rest frame for it.
1.5
a) The formula for length contraction (or Lorentz contraction) in special relativity is given by
where ℓ\ell is the length of an object in its rest frame O\mathcal{O} and ℓ^(')\ell^{\prime} is the corresponding length of the object in a frame O^(')\mathcal{O}^{\prime} moving with constant velocity v(0 <= v < c)v(0 \leq v<c) relative to the rest frame O\mathcal{O} of the object. Note that 0 < ℓ^(') <= ℓ0<\ell^{\prime} \leq \ell, since 1 <= gamma(v) < oo1 \leq \gamma(v)<\infty, so the object appears to be shortened in the frame O^(')\mathcal{O}^{\prime} moving relative to the rest frame O\mathcal{O}.
The length contraction formula can be derived using the fact that the length of the object in a given frame is the spatial distance between the ends of the object measured at the same time in that frame. Consider the time t^(')t^{\prime} in S^(')S^{\prime} to be the time at which we measure the position of the ends of the object, which are then separated by ℓ^(')\ell^{\prime}. Calling the events at the ends of the object AA and BB, respectively, the spatial distance between them in S^(')S^{\prime} is therefore ℓ^(')\ell^{\prime} and the temporal separation is zero. At the same time, the spatial and temporal separations between the events in SS are ℓ\ell and Delta t\Delta t, respectively, as the object is at rest in SS. For the events to be simultaneous in S^(')S^{\prime}, the Lorentz transformation requires
where tt is the time in the frame O\mathcal{O} and t^(')t^{\prime} is the corresponding time in the frame O^(')\mathcal{O}^{\prime} moving with constant velocity v(0 <= v < c)v(0 \leq v<c) relative to the frame O\mathcal{O}. Note that t > t^(')t>t^{\prime}, since 1 <= gamma(v) < oo1 \leq \gamma(v)<\infty, so an observer in the frame O\mathcal{O} measures a longer time than an observer in O^(')\mathcal{O}^{\prime} measures between two events which occur at the same spatial point in the frame O^(')\mathcal{O}^{\prime}.
In order to derive the time dilation formula, we consider the worldline of an observer at rest in S^(')S^{\prime}. In S^(')S^{\prime}, the spatial and temporal differences between two events on the worldline are given by zero and t^(')t^{\prime}, respectively. In the frame SS, the observer is instead moving at speed vv and the spatial and temporal separations are therefore tt and vtv t, respectively. As the spacetime interval between the events is invariant, we find that
where a=1//sqrt2ma=1 / \sqrt{2} \mathrm{~m}. Note that there is no length contraction in the yy-direction. vec(v)\vec{v}
Figure 3.1 A rod inclined 45^(@)45^{\circ} in the xyx y-plane with respect to the xx-axis and the angle theta\theta in the x^(')y^(')x^{\prime} y^{\prime}-plane with respect to the x^(')x^{\prime}-axis. The relative speed between the two planes is vv.
in the rest frame of the Earth, which explains that many muons arrive at the surface.
This can also be derived in the muons' rest frame: Due to length contraction, the thickness of the atmosphere (10km)(10 \mathrm{~km}) in the rest frame of the Earth will correspond to 10^(4)//22m~~450m10^{4} / 22 \mathrm{~m} \approx 450 \mathrm{~m} in the rest frame of the muon. During the lifetime of the muon, the Earth will move the distance tau_(0)v=660*0.999m~~660m\tau_{0} v=660 \cdot 0.999 \mathrm{~m} \approx 660 \mathrm{~m} in the rest frame of the muon, which is larger than 450 m .
1.8
Introduce the rest frames of the station and the train as KK and K^(')K^{\prime}, respectively.
a) For the two events that involve the markings on the track, we have
Let the coordinates of the front and rear end of the express space cruiser be x_(F)=x_{F}=(-tau c,L,0,0)(-\tau c, L, 0,0) and x_(R)=(0,0,0,0)x_{R}=(0,0,0,0), respectively, in the rest frame of the cruiser at the time when the rear watchman sees the rear light go on. The difference between these are Delta x=(-tau c,L,0,0)\Delta x=(-\tau c, L, 0,0). In the rest frame of the hitchhiker, this difference is given by an inverse Lorentz transformation to the rest frame of the asteroid with velocity -v,v-v, v being the velocity of the cruiser in the rest frame of the asteroid. His/her time difference for the lightening of the lanterns is then Deltax^('0)//c=-tau cosh theta+\Delta x^{\prime 0} / c=-\tau \cosh \theta+(L//c)sinh theta(L / c) \sinh \theta. However, this time difference is equal to zero. Thus, we have
Using the fact that the interval between two points in spacetime is invariant, i.e., Deltas^(2)=Deltas^('2)\Delta s^{2}=\Delta s^{\prime 2} or c^(2)Deltat^(2)-Deltax^(2)=c^(2)Deltat^('2)-Deltax^('2)c^{2} \Delta t^{2}-\Delta x^{2}=c^{2} \Delta t^{\prime 2}-\Delta \boldsymbol{x}^{\prime 2}, gives together with the information in the problem text
Now, using the length contraction formula ℓ^(')=ℓgamma(v)\ell^{\prime}=\ell \gamma(v), where gamma(v)=(1)/(sqrt(1-v^(2)//c^(2)))\gamma(v)=\frac{1}{\sqrt{1-v^{2} / c^{2}}}, one obtains
Lorentz contraction only takes place for the projection of the rod that lies along the xx-axis. This projection is ℓcos theta\ell \cos \theta. The orthogonal component is ℓsin theta\ell \sin \theta. Lorentz contraction of the xx-component is then ℓcos thetasqrt(1-v^(2)//c^(2))\ell \cos \theta \sqrt{1-v^{2} / c^{2}}. Therefore, to the moving observer, the length of the rod is
Thus, the answer is ℓ^(')=ℓsqrt(1-((v)/(c))^(2)cos^(2)theta)\ell^{\prime}=\ell \sqrt{1-\left(\frac{v}{c}\right)^{2} \cos ^{2} \theta}.
1.12
The spacetime interval of the two events AA and BB with coordinates x_(A)x_{A} and x_(B)x_{B} is s=(x_(A)-x_(B))^(2)s=\left(x_{A}-x_{B}\right)^{2}. For observer KK in SS, this is s=-L^(2)s=-L^{2}. In the rest frame of K^(')K^{\prime} we have s=c^(2)Deltat^('2)-L^('2)s=c^{2} \Delta t^{\prime 2}-L^{\prime 2}. Therefore, we obtain L=sqrt(L^('2)-c^(2)Deltat^('2))L=\sqrt{L^{\prime 2}-c^{2} \Delta t^{\prime 2}}. Thus, the answer is L=sqrt(L^('2)-c^(2)Deltat^('2))L=\sqrt{L^{\prime 2}-c^{2} \Delta t^{\prime 2}}.
1.13
The invariant interval is (x_(alpha)-x_(beta))^(2)=(x_(alpha)^(')-x_(beta)^('))^(2)\left(x_{\alpha}-x_{\beta}\right)^{2}=\left(x_{\alpha}^{\prime}-x_{\beta}^{\prime}\right)^{2}.
a) Denote the distance to the beta\beta event for S^(')S^{\prime} by L^(')L^{\prime}. We set c=1c=1 and calculate the length in ly. Then, we have
b) The Lorentz transformation from KK to K^(')K^{\prime} gives
{:[(3.43)1=2gamma-10 v gamma],[(3.44)L^(')//ly=-2v gamma+10 gamma]:}\begin{align*}
1 & =2 \gamma-10 v \gamma \tag{3.43}\\
L^{\prime} / \mathrm{ly} & =-2 v \gamma+10 \gamma \tag{3.44}
\end{align*}
where gamma=1//sqrt(1-v^(2))\gamma=1 / \sqrt{1-v^{2}}. Solving this equation yields v~~0.1v \approx 0.1 as the permissible root.
1.14
The smallest energy required for a muon to hit the ground within its mean life in the rest frame of the Earth is given when the direction of the muon is vertical. In this case, the muon must travel the distance ℓ=10km\ell=10 \mathrm{~km} in the time tau\tau, which is the time dilated mean life tau_(0)\tau_{0} of the muon in its rest frame. It follows that
{:(3.45)ℓ <= v tau=v gamma(v)tau_(0):}\begin{equation*}
\ell \leq v \tau=v \gamma(v) \tau_{0} \tag{3.45}
\end{equation*}
The velocity of the muon is given by v=pc^(2)//Ev=p c^{2} / E, and thus, we have
Inserting the numerical values, we find that pc≳15 mc^(2)≃1.6GeVp c \gtrsim 15 m c^{2} \simeq 1.6 \mathrm{GeV}, and thus, it holds that E≃pc≳1.6GeVE \simeq p c \gtrsim 1.6 \mathrm{GeV}.
1.15
A time interval for a muon and a time interval for an observer in the lab frame are related through the time dilation formula
{:(3.48)gamma(v)d tau=dt:}\begin{equation*}
\gamma(v) d \tau=d t \tag{3.48}
\end{equation*}
where d taud \tau is the time interval for the muon, dtd t is the time interval for the observer in the lab frame, and vv is the muon velocity. The muon velocity is constant (since the total energy is constant) and given by
The average lifetime of the muon in the lab frame is therefore 22 mus22 \mu \mathrm{~s}. Since the muon is highly relativistic (E≫m)(E \gg m), the length traveled by the muon in the lab frame in the average lifetime is given by
{:(3.52)ℓ=vt≃ct≃3*10^(8)m//s*22*10^(-6)s=6600m.:}\begin{equation*}
\ell=v t \simeq c t \simeq 3 \cdot 10^{8} \mathrm{~m} / \mathrm{s} \cdot 22 \cdot 10^{-6} \mathrm{~s}=6600 \mathrm{~m} . \tag{3.52}
\end{equation*}
The circumference of the circular accelerator is given by
{:(3.53)L=2pi r≃300m:}\begin{equation*}
L=2 \pi r \simeq 300 \mathrm{~m} \tag{3.53}
\end{equation*}
Thus, the average number of turns taken by a muon is given by
where A=6A=6 is the area in KK.
b) Let hh be the height of the triangle in K^(')K^{\prime}. Since A=6=ch//2A=6=c h / 2, we get h=h=12//512 / 5. We have c^(')=c//gamma,h^(')=hc^{\prime}=c / \gamma, h^{\prime}=h, and thus, the area becomes A^(')=c^(')h^(')//2=A//gammaA^{\prime}=c^{\prime} h^{\prime} / 2=A / \gamma as in a). Let c_(1)c_{1} be the distance in KK from the corner, where the aa - and cc-sides of the triangle meet in the point, where the height intersects the cc-side in a right angle. Obviously, c_(1)=sqrt(a^(2)-h^(2))c_{1}=\sqrt{a^{2}-h^{2}}. In K^(')K^{\prime}, the same distance is c_(1)^(')=c_(1)//gammac_{1}^{\prime}=c_{1} / \gamma, and thus, we obtain
The rest frame of the pole is denoted SS with coordinates X=(t,x,y,z)X=(t, x, y, z), where the pole has length LL. The lab frame is denoted S^(')S^{\prime} with coordinates X^(')=(t^('),x^('),y^('),z^('))X^{\prime}=\left(t^{\prime}, x^{\prime}, y^{\prime}, z^{\prime}\right), in which the pole length is L^(')=L//gammaL^{\prime}=L / \gamma, where gamma=1//sqrt(1-v^(2)//c^(2))\gamma=1 / \sqrt{1-v^{2} / c^{2}}. In the lab frame, the pole is moving in the negative xx-direction with speed vv, parallel to the xx-axis at some distance yy. Everything occurs in the spatial plane z=0z=0, so we can ignore the zz-direction. Denote the events at which light is emitted from front of the pole by A, from the mark on the pole by B, and from the end of the pole by C. The three rays of light all reach the origin event in S^(')S^{\prime}, i.e., the spatial origin at t^(')=0t^{\prime}=0. Since the three light rays are emitted from different positions, but all reach the origin at the same time, the light rays were emitted at different times t^(')t^{\prime}. We now consider each of the three light rays separately.
Event A: Let X_(A)^(')=(t^('),x^('),y^('))X_{\mathrm{A}}^{\prime}=\left(t^{\prime}, x^{\prime}, y^{\prime}\right) denote the coordinates in S^(')S^{\prime} of the event A. Since the line element along the 4-path of a light signal vanishes, we have
Defining the time at which the event A occurs to be t^(')=-delta_(A)t^{\prime}=-\delta_{\mathrm{A}}, we obtain the coordinates of A in S^(')S^{\prime} as
Event B: Let X_(B)^(')=(t^('),x^('),y^('))X_{\mathrm{B}}^{\prime}=\left(t^{\prime}, x^{\prime}, y^{\prime}\right) denote the coordinates in S^(')S^{\prime} of the event B. In this case, the angle is pi//4\pi / 4, so we have
Event C: Let X_(C)^(')=(t^('),x^('),y^('))X_{\mathrm{C}}^{\prime}=\left(t^{\prime}, x^{\prime}, y^{\prime}\right) denote the coordinates in S^(')S^{\prime} of the event C. Similar to events A and B , defining the time of the event C as t^(')=-delta_(C)t^{\prime}=-\delta_{\mathrm{C}}, we obtain
The event A occurs at the front of the pole at x^(')=delta_(A)//2x^{\prime}=\delta_{\mathrm{A}} / 2 at time -delta_(A)-\delta_{\mathrm{A}}, whereas the event C occurs at the end of the pole at x^(')=3delta_(A)//2x^{\prime}=3 \delta_{\mathrm{A}} / 2 at time -sqrt3delta_(A)-\sqrt{3} \delta_{\mathrm{A}}. The pole moves at speed vv in the negative xx-direction. Therefore, at time -delta_(A)-\delta_{\mathrm{A}}, the end of the pole is located at
In S^(')S^{\prime}, we find that the length of the pole is the difference between this and x_(A)^(')=delta_(A)//2x_{\mathrm{A}}^{\prime}=\delta_{\mathrm{A}} / 2, which occurs simultaneously at the front, so
However, in S^(')S^{\prime}, we know that the length of the pole is L^(')=L//gammaL^{\prime}=L / \gamma, which means that we have determined delta_(A)\delta_{\mathrm{A}} to be
and therefore, we have determined all coordinates of the events A,B\mathrm{A}, \mathrm{B}, and C in S^(')S^{\prime}.
Now, we determine the distance between the front of the pole and the mark on the pole. At time -delta_(A)-\delta_{\mathrm{A}}, the x^(')x^{\prime}-coordinate of the mark is given by
Therefore, in S^(')S^{\prime}, the ratio rr between the distance from the front of the pole to the mark on the pole and the length of the whole pole is
Note that rr is the same in all inertial frames due to the linearity of Lorentz transformations. For example, in the rest frame of a pole with length LL and the front at the origin, the mark would be located at the xx-coordinate rLr L.
1.18
a) In the initial rest frame of the spaceships, they are at a constant distance LL from each other during the entire accelerating phase. However, this length is the length contraction of the string in the new rest frame and the proper length will then be L^(')=gamma LL^{\prime}=\gamma L. Trying to force the string to be of length LL in the initial rest frame will therefore break it.
b) The initial distance between the two spaceships is 40 km . The spaceships accelerate with a constant acceleration a=1//50c//sa=1 / 50 \mathrm{c} / \mathrm{s}. When the spaceships stop after 30 s in the initial rest frame, the speed of the spaceships will then be v=at=3c//5v=a t=3 c / 5. Hence, the gamma factor can be computed to be
Thus, the distance between the two spaceships is L^(')=gamma L=5//4*40km=50kmL^{\prime}=\gamma L=5 / 4 \cdot 40 \mathrm{~km}=50 \mathrm{~km}.
1.19
Spacetime diagrams showing the events described in the problem are shown in Figure 3.2. Part (a) of the diagram depicts the events using the time and xx axes of the rest frame of Professor Einstein as orthogonal, while part (b) is the same spacetime diagram with the axes of Professor Wolf orthogonal. The coordinate axes of Professor Einstein are marked by t_(E)t_{\mathrm{E}} and x_(E)x_{\mathrm{E}}, respectively, while those of Professor Wolf are marked by t_(W)t_{\mathrm{W}} and x_(W)x_{\mathrm{W}}, respectively. The light gray shaded region is the train, the thick black lines and the thick dark gray lines represent the worldlines of Professor Einstein and Professor Wolf, respectively, while the thick white lines represent the light signals.
Figure 3.2 Spacetime diagrams describing the events of Problem 1.19.
From the diagrams it is clear that
a) The light signals will always reach Professor Einstein at the same time. This is something physically observable and cannot depend on the set of coordinates used to describe the events. (In fact, a light signal reaching Professor Einstein is an event and both light signals reach him at the same event).
b) Professor Wolf sees the light signals being reflected at different times. This can be seen as the line between the reflection points is not parallel to the spatial axis in Professor Wolf's rest frame, i.e., it is not a simultaneity in that frame.
c) The signals will reach Professor Wolf at different times. The signal from the back of the train has already reached Professor Wolf before it reaches Professor Einstein, while the signal from the front reflection will reach him after it reaches Professor Einstein.
1.20
Let us consider the situation seen in the frame SS where the two spaceships have equal but opposite velocities. The situation can then be described through the spacetime diagram in Figure 3.3. Since the velocities of both spaceships relative to this frame are the same, they are equally Lorentz contracted and the ship with rest length 2L2 L is twice as long (i.e., light gray shaded region) as the ship with rest length LL (i.e., dark gray shaded region). The coordinates have been chosen such that AA occurs in the spacetime origin and the event BB has been marked in the diagram. The time axes of both the frame S^(')S^{\prime}, in which the shorter ship is at rest, and S^('')S^{\prime \prime}, in which the longer is at rest, are also shown. The dark (light) gray line represents the surface simultaneous with BB in S^(')(S^(''))S^{\prime}\left(S^{\prime \prime}\right) and the intersection with the t^(')(t^(''))t^{\prime}\left(t^{\prime \prime}\right)-axis has been marked. Since the t^(')t^{\prime} and t^('')t^{\prime \prime} axes are tilted with the same angle relative to the tt-axis,
Figure 3.3 The spacetime diagram describing the situation in Problem 1.20.
Figure 3.4 Spacetime diagrams describing the situation in Problem 1.21.
they both have the same normalization. Thus, t_(2L) > t_(L)t_{2 L}>t_{L}, i.e., the time measured in the rest frame of the shorter rocket is shorter.
1.21
The situation is described by the spacetime diagrams in Figure 3.4. In part (a) of the diagram, we see the situation described in SS, the rest frame of the Earth. The light gray thick line (marked by mu\mu ) corresponds to the muon worldline, the length of which is the muon eigentime to reach the Earth's surface. The black solid curve is a hyperbolic curve, which means that straight worldlines from the origin to that curve will have the same length, i.e., eigentime. From this we draw the conclusion
that the eigentime experienced by the muons will be significantly smaller than the time taken to perform the travel from the top of the atmosphere to the surface and the muons thus will not have sufficient time to decay. More quantitative statements cannot be made without further information. The t^(')t^{\prime} - and x^(')x^{\prime}-axes of S^(')S^{\prime}, the muon rest frame, are also shown along with the light gray thin line, which shows the light cone. For completeness, we also show how the situation looks in S^(')S^{\prime} in part (b) of the figure.
1.22
The problem is most easily solved by considering the two events when the ends of the blade reaches the y^(')=y=0y^{\prime}=y=0 plane. Letting the tip of the guillotine cutting the plane be the origin x=x^(')=t=t^(')=0x=x^{\prime}=t=t^{\prime}=0, the other end of the guillotine cutting the plane will have the coordinates
where we have taken into account that the blade is length contracted in the yy direction in SS with a factor gamma_(u)=1//sqrt(1-u^(2))\gamma_{u}=1 / \sqrt{1-u^{2}}. Transforming this to the S^(')S^{\prime} frame, we find that
If the blade edge is horizontal in S^(')S^{\prime}, then all points on the edge will cut the y^(')=0y^{\prime}=0 plane at the same time and since t^(')=0t^{\prime}=0 for the point cutting it, we find that
The spacetime diagram in Figure 3.5 shows the worldlines of the observer, the two lights, and a light signal from the lights that arrives to the observer at the same time.
The positions where the lights were when this signal was emitted will be the positions seen by the observer at t=t^(**)t=t^{*} and the seen separation LL will be given by the difference of these positions.
The worldlines of the lights are given by
{:(3.79)x_(1)(t)=x_(1,0)-vt","quadx_(2)(t)=x_(2,0)-vt:}\begin{equation*}
x_{1}(t)=x_{1,0}-v t, \quad x_{2}(t)=x_{2,0}-v t \tag{3.79}
\end{equation*}
If the time coordinate is chosen such that the signal is emitted from x_(2)(0)x_{2}(0) at t=0t=0, the signal worldline is given by
Figure 3.5 Spacetime diagram showing the worldlines of an observer, two lights (L1 and L2), and a light signal arriving to the observer at t=t^(**)t=t^{*}.
and intersects the worldlines of light 2 at x_(2,0)x_{2,0}. The intersection of the worldline of light 1 is given by
where we have used that x_(2,0)-x_(1,0)=(ℓ_(0))/(gamma)x_{2,0}-x_{1,0}=\frac{\ell_{0}}{\gamma} by Lorentz contraction. The seen distance between the lights is therefore
An object traveling at speed cc in the frame SS has a position given by x=cnt+x_(0)\boldsymbol{x}=c \boldsymbol{n} t+\boldsymbol{x}_{0}, where n^(2)=1\boldsymbol{n}^{2}=1. The Lorentz transform then becomes
The velocity in the new inertial frame S^(')S^{\prime} is given by
{:(3.85)v^(')=(dx^('))/(dt^('))=(dx^(')//dt)/(dt^(')//dt)=((gamma(cn_(1)-v),cn_(2),cn_(3)))/(gamma(1-vn_(1)//c)):}\begin{equation*}
\boldsymbol{v}^{\prime}=\frac{d \boldsymbol{x}^{\prime}}{d t^{\prime}}=\frac{d \boldsymbol{x}^{\prime} / d t}{d t^{\prime} / d t}=\frac{\left(\gamma\left(c n_{1}-v\right), c n_{2}, c n_{3}\right)}{\gamma\left(1-v n_{1} / c\right)} \tag{3.85}
\end{equation*}
Squaring this relation leads to
{:[v^('2)=(gamma^(2)(c^(2)n_(1)^(2)-2cn_(1)v+v^(2))+c^(2)(n_(2)^(2)+n_(3)^(2)))/(gamma^(2)(1-2vn_(1)//c+v^(2)n_(1)^(2)//c^(2)))],[=(gamma^(2)[c^(2)n_(1)^(2)-2cn_(1)v+v^(2)+(c^(2)-v^(2))(n_(2)^(2)+n_(3)^(2))])/(gamma^(2)(1-2vn_(1)//c+v^(2)n_(1)^(2)//c^(2)))],[(3.86)=(c^(2)-2cn_(1)v+v^(2)n_(1)^(2))/(1-2vn_(1)//c+v^(2)n_(1)^(2)//c^(2))=c^(2)","]:}\begin{align*}
\boldsymbol{v}^{\prime 2} & =\frac{\gamma^{2}\left(c^{2} n_{1}^{2}-2 c n_{1} v+v^{2}\right)+c^{2}\left(n_{2}^{2}+n_{3}^{2}\right)}{\gamma^{2}\left(1-2 v n_{1} / c+v^{2} n_{1}^{2} / c^{2}\right)} \\
& =\frac{\gamma^{2}\left[c^{2} n_{1}^{2}-2 c n_{1} v+v^{2}+\left(c^{2}-v^{2}\right)\left(n_{2}^{2}+n_{3}^{2}\right)\right]}{\gamma^{2}\left(1-2 v n_{1} / c+v^{2} n_{1}^{2} / c^{2}\right)} \\
& =\frac{c^{2}-2 c n_{1} v+v^{2} n_{1}^{2}}{1-2 v n_{1} / c+v^{2} n_{1}^{2} / c^{2}}=c^{2}, \tag{3.86}
\end{align*}
where we have used that n_(1)^(2)+n_(2)^(2)+n_(3)^(2)=1n_{1}^{2}+n_{2}^{2}+n_{3}^{2}=1. Thus, the object travels at speed cc also in S^(')S^{\prime}.
1.25
Let FF and RR denoted the front and the rear of the train, respectively. In the rest frame of the train at the time of passing of the rear in front of the station man, we take the coordinates to be x_(F)^(')=(0,L)x_{F}^{\prime}=(0, L) and x_(R)^(')=(0,0)x_{R}^{\prime}=(0,0). For the stationman, these coordinates are instead x_(F)=(x_(F)^(0),x_(F)^(1))x_{F}=\left(x_{F}^{0}, x_{F}^{1}\right) and x_(R)=(x_(R)^(0),x_(R)^(1))x_{R}=\left(x_{R}^{0}, x_{R}^{1}\right), which are obtained from the first ones by means of an inverse Lorentz transformation along the xx-axis, the direction of motion of the train, given by
{:(3.87){[x_(i)^(0)=x_(i)^('0)cosh theta+x_(i)^('1)sinh theta],[x_(i)^(1)=x_(i)^('0)sinh theta+x_(i)^('')cosh theta],quad" where "i=F,R" and "tanh theta=(v)/(c):}:}\left\{\begin{array}{l}
x_{i}^{0}=x_{i}^{\prime 0} \cosh \theta+x_{i}^{\prime 1} \sinh \theta \tag{3.87}\\
x_{i}^{1}=x_{i}^{\prime 0} \sinh \theta+x_{i}^{\prime \prime} \cosh \theta
\end{array}, \quad \text { where } i=F, R \text { and } \tanh \theta=\frac{v}{c}\right.
Let AA and BB be the front and rear end of the train, respectively. Since the velocity of light is cc for all observers, the times are given by t_(1)=-Lx//ct_{1}=-L x / c and t_(2)=-L(1-x)//ct_{2}=-L(1-x) / c. These times are related to the times t_(1)^(')t_{1}^{\prime} and t_(2)^(')t_{2}^{\prime} determined by the observer on the ground by a Lorentz transformation
{:[(3.90)ct_(2)^(')=ct_(2)cosh theta+[-L(1-x)]sinh theta],[(3.91)ct_(1)^(')=ct_(1)cosh theta+xL sinh theta]:}\begin{align*}
& c t_{2}^{\prime}=c t_{2} \cosh \theta+[-L(1-x)] \sinh \theta \tag{3.90}\\
& c t_{1}^{\prime}=c t_{1} \cosh \theta+x L \sinh \theta \tag{3.91}
\end{align*}
where we have put the origin at O=O^(')O=O^{\prime}. This gives c(t_(2)^(')-t_(1)^('))=c(t_(2)-t_(1))cosh theta-c\left(t_{2}^{\prime}-t_{1}^{\prime}\right)=c\left(t_{2}-t_{1}\right) \cosh \theta-L sinh thetaL \sinh \theta. If t_(2)^(')-t_(1)^(')=0t_{2}^{\prime}-t_{1}^{\prime}=0, we obtain v=c tanh theta=(2x-1)cv=c \tanh \theta=(2 x-1) c, and v=0v=0 therefore
implies x=1//2x=1 / 2. This means that if x < 1//2x<1 / 2 the train has to move in opposite direction to when x > 1//2x>1 / 2, for the situation to occur, i.e., AA and BB change roles of being rear and front, respectively.
One can also calculate the invariant interval s^(2)=c^(2)(t_(2)-t_(1))^(2)-L^(2)=c^(2)(t_(2)^(')-:}s^{2}=c^{2}\left(t_{2}-t_{1}\right)^{2}-L^{2}=c^{2}\left(t_{2}^{\prime}-\right.t_(1)^('))^(2)-L^('2)\left.t_{1}^{\prime}\right)^{2}-L^{\prime 2} in the two frames and use the length contraction formula to obtain the velocity. In this treatment, the sign of the velocity must be discussed separately.
1.27
a) Suppose the particle moves through the origin of KK. Then, the event A=A= ( ct,0,0,-utc t, 0,0,-u t ) belongs to the worldline of the particle. Transforming AA to K^(')K^{\prime} using the standard Lorentz transformation, we find that A^(')=(ct gamma(v),-vt gamma(v),0,-ut)A^{\prime}=(\operatorname{ct\gamma }(v),-v t \gamma(v), 0,-u t). The angle is therefore given by
The observer in K^(')K^{\prime} measures the triangle at time x^('0)=0x^{\prime 0}=0. Using Eq. (3.97) together with x^('0)=0x^{\prime 0}=0, implies that x^(0)=(v)/(c)x^(1)x^{0}=\frac{v}{c} x^{1}. Inserting x^(0)=(v)/(c)x^(1)x^{0}=\frac{v}{c} x^{1} into Eq. (3.98), yields
i.e., x^('1)=(1)/(gamma(v))x^(1)x^{\prime 1}=\frac{1}{\gamma(v)} x^{1}, which is the Lorentz length contraction formula.
In KK :
All three sides of the triangle have length ℓ\ell and all three angles in the triangle are 60^(@)60^{\circ} (i.e., (pi)/(3)\frac{\pi}{3} ). One of the sides in the triangle (the base b=ℓb=\ell ) is parallel to the x^(1)x^{1}-axis. Using Pythagoras' theorem, ℓ^(2)=((b)/(2))^(2)+h^(2)\ell^{2}=\left(\frac{b}{2}\right)^{2}+h^{2}, one finds the length of the altitude (the height) of the triangle to be h=(sqrt3)/(2)*ℓh=\frac{\sqrt{3}}{2} \cdot \ell.
In K^(')K^{\prime} :
The length of base of the triangle is: b^(')=(1)/(gamma(v))*b=(1)/(gamma(v))ℓb^{\prime}=\frac{1}{\gamma(v)} \cdot b=\frac{1}{\gamma(v)} \ell. The length of the altitude (the height) of the triangle is: h^(')=h=(sqrt3)/(2)ℓh^{\prime}=h=\frac{\sqrt{3}}{2} \ell [using Eq. (3.99)]. Assume that the length of the two other sides of the triangle is ℓ^(')\ell^{\prime}. Again, using Pythagoras' theorem, ℓ^('2)=((b^('))/(2))^(2)+h^('2)\ell^{\prime 2}=\left(\frac{b^{\prime}}{2}\right)^{2}+h^{\prime 2}, one finds the length of the two other sides as
The base angle alpha\alpha can be obtained from the relation ℓ^(')*cos alpha=(b^('))/(2)\ell^{\prime} \cdot \cos \alpha=\frac{b^{\prime}}{2} and the apex angle beta\beta from the relation ℓ^(')*sin((beta)/(2))=(b^('))/(2)\ell^{\prime} \cdot \sin \frac{\beta}{2}=\frac{b^{\prime}}{2}. The results are: alpha=arccos((1)/(sqrt(1+3gamma(v)^(2))))\alpha=\arccos \frac{1}{\sqrt{1+3 \gamma(v)^{2}}} and beta=2arcsin((1)/(sqrt(1+3gamma(v)^(2))))\beta=2 \arcsin \frac{1}{\sqrt{1+3 \gamma(v)^{2}}}.
1.30
For two events, E_(1)E_{1} and E_(2)E_{2} that occur at the left and right endpoints of the rod, respectively, we have
Let the axis of the cylinder coincide with the xx-axis in KK. The straight line on the cylinder surface is described by the equation
{:(3.105)varphi=omega t:}\begin{equation*}
\varphi=\omega t \tag{3.105}
\end{equation*}
where varphi\varphi is the angle of rotation around the xx-axis. We now transform the equation varphi=omega t\varphi=\omega t to K^(')K^{\prime}, that moves with velocity vv in the xx-direction. We find that
{:(3.106)varphi=varphi^(')quad" and "quad t=gamma(v)(t^(')+(vx^('))/(c^(2))):}\begin{equation*}
\varphi=\varphi^{\prime} \quad \text { and } \quad t=\gamma(v)\left(t^{\prime}+\frac{v x^{\prime}}{c^{2}}\right) \tag{3.106}
\end{equation*}
which means that the equation of motion of the straight line relative to K^(')K^{\prime} is described by the equation
so that to the observer in K^(')K^{\prime}, the straight line appears as a twisted line around the cylinder.
1.32
Let E_(1)E_{1} and E_(2)E_{2} be the events of turning on two compartment lights. If KK and K^(')K^{\prime} are the rest frames of the station and the train, respectively, then
However, it holds that Delta x=u Delta t\Delta x=u \Delta t, which gives u=c^(2)//vu=c^{2} / v.
1.33
Let KK and K^(')K^{\prime} be the rest frames of the star and spaceship, respectively. Furthermore, let the planet have its orbit in the xyx y-plane in the coordinate system KK,
i.e., z=0z=0 for the planet. The spacetime trajectory of the planet in KK is then x=(x^(0),x)=(ct,x)x=\left(x^{0}, \boldsymbol{x}\right)=(c t, \boldsymbol{x}), where x=(R cos omega t,R sin omega t,0)\boldsymbol{x}=(R \cos \omega t, R \sin \omega t, 0). The orbit of the planet in K^(')K^{\prime} is now given by the Lorentz transformation:
where gamma(v)=(1)/(sqrt(1-v^(2)//c^(2)))\gamma(v)=\frac{1}{\sqrt{1-v^{2} / c^{2}}}. The spaceship is moving along the positive zz-axis with velocity vv. Therefore, we have
{:[(3.111)t^(')=gamma(v)t],[(3.112)x^(')=R cos omega t=R cos omega(t^('))/(gamma(v))=R cos omega^(')t^(')],[(3.113)y^(')=R sin omega t=R sin omega(t^('))/(gamma(v))=R sin omega^(')t^(')],[(3.114)z^(')=-gamma(v)vt=-vt^(')]:}\begin{align*}
& t^{\prime}=\gamma(v) t \tag{3.111}\\
& x^{\prime}=R \cos \omega t=R \cos \omega \frac{t^{\prime}}{\gamma(v)}=R \cos \omega^{\prime} t^{\prime} \tag{3.112}\\
& y^{\prime}=R \sin \omega t=R \sin \omega \frac{t^{\prime}}{\gamma(v)}=R \sin \omega^{\prime} t^{\prime} \tag{3.113}\\
& z^{\prime}=-\gamma(v) v t=-v t^{\prime} \tag{3.114}
\end{align*}
where omega^(')=omega//gamma(v)\omega^{\prime}=\omega / \gamma(v). Thus, the spacetime trajectory of the planet in K^(')K^{\prime} is given by x^(')=(x^('0),x^('))=(ct^('),x^('))x^{\prime}=\left(x^{\prime 0}, \boldsymbol{x}^{\prime}\right)=\left(c t^{\prime}, \boldsymbol{x}^{\prime}\right), where
where tanh theta=(v)/(c)\tanh \theta=\frac{v}{c}, and the Lorentz transformation between BB and CC ( CC is moving with velocity v^(')v^{\prime} along the positive x^('2)x^{\prime 2}-axis in K^(')K^{\prime} ) is given by
where tanh theta^(')=(v^('))/(c)\tanh \theta^{\prime}=\frac{v^{\prime}}{c}.
Inserting the equations for x^('0)x^{\prime 0} and x^('2)x^{\prime 2} into the equation for x^(''0)x^{\prime \prime 0}, we obtain
where tanh theta^('')-=(v^(''))/(c)\tanh \theta^{\prime \prime} \equiv \frac{v^{\prime \prime}}{c}. The velocity v^('')v^{\prime \prime} is the (magnitude of) the relative velocity between AA and CC. Using the hint, this means that
We know that gamma(v^(''))=cosh theta^('')\gamma\left(v^{\prime \prime}\right)=\cosh \theta^{\prime \prime}. It follows that gamma(v^(''))=cosh theta cosh theta^(')=\gamma\left(v^{\prime \prime}\right)=\cosh \theta \cosh \theta^{\prime}=gamma(v)gamma(v^('))\gamma(v) \gamma\left(v^{\prime}\right), and thus, the time dilation formula between the time intervals Delta t-=\Delta t \equivt_(E_(2))-t_(E_(1))t_{E_{2}}-t_{E_{1}} and Deltat^('')-=t_(E_(2))^('')-t_(E_(1))^('')\Delta t^{\prime \prime} \equiv t_{E_{2}}^{\prime \prime}-t_{E_{1}}^{\prime \prime} is given by
However, x^(2)=0x^{2}=0 implies that (x^(0))^(2)=(x^(1))^(2)+(x^(2))^(2)+(x^(3))^(2)\left(x^{0}\right)^{2}=\left(x^{1}\right)^{2}+\left(x^{2}\right)^{2}+\left(x^{3}\right)^{2}, so we can write (x^(1))^(2)+(x^(2))^(2)=(x^(0))^(2)-(x^(3))^(2)=(x^(0)-x^(3))(x^(0)+x^(3))\left(x^{1}\right)^{2}+\left(x^{2}\right)^{2}=\left(x^{0}\right)^{2}-\left(x^{3}\right)^{2}=\left(x^{0}-x^{3}\right)\left(x^{0}+x^{3}\right). Therefore, we obtain
A Lorentz transformation is linear and we will derive the expressions for a boost in the xx-direction. The general linear transformation is given by
{:[(3.134)x^(')=gamma x+bt],[(3.135)t^(')=Ax+Bt]:}\begin{align*}
x^{\prime} & =\gamma x+b t \tag{3.134}\\
t^{\prime} & =A x+B t \tag{3.135}
\end{align*}
for a transformation from an inertial frame SS to an inertial frame S^(')S^{\prime} moving with a speed vv relative to SS. In the case that x=vtx=v t, i.e., moving along with S^(')S^{\prime}, we must have x^(')=0x^{\prime}=0. Hence, we have
{:(3.136)x^(')=gamma vt+bt=(gamma v+b)t=0quad=>quad b=-gamma v:}\begin{equation*}
x^{\prime}=\gamma v t+b t=(\gamma v+b) t=0 \quad \Rightarrow \quad b=-\gamma v \tag{3.136}
\end{equation*}
where for the second equation we have assumed that the situation must be symmetric to the first one. Furthermore, we must have t=x//ct=x / c and t^(')=x^(')//ct^{\prime}=x^{\prime} / c for a light ray. Therefore, we find that
{:[(3.139)x^(')=gamma(1-(v)/(c))x],[(3.140)x=gamma(1+(v)/(c))x^(')]:}\begin{align*}
x^{\prime} & =\gamma\left(1-\frac{v}{c}\right) x \tag{3.139}\\
x & =\gamma\left(1+\frac{v}{c}\right) x^{\prime} \tag{3.140}
\end{align*}
Thus, inserting the first equation into the second one, we obtain
{:(3.141)x=gamma^(2)(1-(v^(2))/(c^(2)))x=>gamma=(1)/(sqrt(1-(v^(2))/(c^(2)))):}\begin{equation*}
x=\gamma^{2}\left(1-\frac{v^{2}}{c^{2}}\right) x \Rightarrow \gamma=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}} \tag{3.141}
\end{equation*}
In addition, we can determine the parameters AA and BB to be A=-gamma(v)/(c^(2))A=-\gamma \frac{v}{c^{2}} and B=gammaB=\gamma by requiring the inverse transformation to take a similar form.
1.37
Calling the events where the light signals reach the rockets 1 and 2 , respectively (with 2 being the event for the rocket traveling with velocity vv ), event 2 is given by the intersection of the worldlines
{:(3.142)x=vt quad" and "quad x=t-t_(0):}\begin{equation*}
x=v t \quad \text { and } \quad x=t-t_{0} \tag{3.142}
\end{equation*}
This leads to
{:(3.143)t_(2)=(t_(0))/(1-v)quad" and "quadx_(2)=vt_(2)=(vt_(0))/(1-v):}\begin{equation*}
t_{2}=\frac{t_{0}}{1-v} \quad \text { and } \quad x_{2}=v t_{2}=\frac{v t_{0}}{1-v} \tag{3.143}
\end{equation*}
By symmetry, t_(1)=t_(2)t_{1}=t_{2} and x_(1)=-x_(2)x_{1}=-x_{2}. It follows that the spacetime separation between the events is given by
{:(3.144)Delta x=x_(2)-x_(1)=(2vt_(0))/(1-v)quad" and "quad Delta t=t_(2)-t_(1)=0:}\begin{equation*}
\Delta x=x_{2}-x_{1}=\frac{2 v t_{0}}{1-v} \quad \text { and } \quad \Delta t=t_{2}-t_{1}=0 \tag{3.144}
\end{equation*}
Lorentz transforming this to the rest frame S^(')S^{\prime} of rocket 2, we find that
a) Relativity of simultaneity means that it is possible for two events that are simultaneous in one inertial frame to not be simultaneous in a different frame. This occurs if there is a spatial separation between the events in the direction of the relative velocity between the frames.
The concept of relativity of simultaneity can be illustrated by the spacetime diagram shown in Figure 3.6. In this spacetime diagram, there are two events A and B , where A occurs at the origin of both frames O\mathcal{O} and O^(')\mathcal{O}^{\prime}, i.e., x_(A)=x_(A)^(')=0x_{\mathrm{A}}=x_{\mathrm{A}}^{\prime}=0, at equal times t_(A)=t_(A)^(')=0t_{\mathrm{A}}=t_{\mathrm{A}}^{\prime}=0, whereas B occurs at x_(B)x_{\mathrm{B}} in O\mathcal{O} at time t_(B)=t_(A)=0t_{\mathrm{B}}=t_{\mathrm{A}}=0, but at x_(B)^(')x_{\mathrm{B}}^{\prime}
Figure 3.6 Relativity of simultaneity: t_(A)=t_(B)t_{\mathrm{A}}=t_{\mathrm{B}}, but t_(A)^(') > t_(B)^(')t_{\mathrm{A}}^{\prime}>t_{\mathrm{B}}^{\prime}.
in O^(')\mathcal{O}^{\prime} at time t_(B)^(')!=t_(A)^(')t_{\mathrm{B}}^{\prime} \neq t_{\mathrm{A}}^{\prime}. In fact, the two events are simultaneous in O\mathcal{O}, since t_(A)=t_(B)t_{\mathrm{A}}=t_{\mathrm{B}}, but they are not simultaneous in O^(')\mathcal{O}^{\prime}, since t_(A)^(') > t_(B)^(')t_{\mathrm{A}}^{\prime}>t_{\mathrm{B}}^{\prime}. In mathematical language, using the Lorentz transformation between O\mathcal{O} and O^(')\mathcal{O}^{\prime} (assuming that the frames are moving with constant speed vv relative to each other), we obtain the relations for B as x_(B)^(')=gamma(v)x_(B)x_{\mathrm{B}}^{\prime}=\gamma(v) x_{\mathrm{B}} and t_(B)^(')=-gamma(v)vx_(B)//c^(2)t_{\mathrm{B}}^{\prime}=-\gamma(v) v x_{\mathrm{B}} / c^{2}, where gamma(v)-=1//sqrt(1-v^(2)//c^(2))\gamma(v) \equiv 1 / \sqrt{1-v^{2} / c^{2}}. Note that x_(B)^(') >= x_(B)x_{\mathrm{B}}^{\prime} \geq x_{\mathrm{B}}, since gamma(v) >= 1\gamma(v) \geq 1, and t_(B)^(') <= 0t_{\mathrm{B}}^{\prime} \leq 0 and directly proportional to the spatial coordinate x_(B)x_{\mathrm{B}}.
b) We compute
and thus, we find that v^(2)=(5at)^(2)\boldsymbol{v}^{2}=(5 a t)^{2}. The condition |v| < c|\boldsymbol{v}|<c implies t_(0) < c//(5a)t_{0}<c /(5 a). We also find
{:(3.148)gamma=(1)/(sqrt(1-(v//c)^(2)))=(1)/(sqrt(1-(5at//c)^(2))):}\begin{equation*}
\gamma=\frac{1}{\sqrt{1-(v / c)^{2}}}=\frac{1}{\sqrt{1-(5 a t / c)^{2}}} \tag{3.148}
\end{equation*}
Thus, we can compute the 4 -velocity and the 4 -acceleration to be
{:[(3.149)(V^(mu))=(c gamma,gamma(dx)/(dt),gamma(dy)/(dt),gamma(dz)/(dt))=gamma(c","3at","4at","0)],[(3.150)(A^(mu))=gamma(d)/(dt)(V^(mu))=gamma(c(d gamma)/(dt),3a(d(gamma t))/(dt),4a(d(gamma t))/(dt),0)=agamma^(4)(25 at//c","3","4","0)]:}\begin{align*}
& \left(V^{\mu}\right)=\left(c \gamma, \gamma \frac{d x}{d t}, \gamma \frac{d y}{d t}, \gamma \frac{d z}{d t}\right)=\gamma(c, 3 a t, 4 a t, 0) \tag{3.149}\\
& \left(A^{\mu}\right)=\gamma \frac{d}{d t}\left(V^{\mu}\right)=\gamma\left(c \frac{d \gamma}{d t}, 3 a \frac{d(\gamma t)}{d t}, 4 a \frac{d(\gamma t)}{d t}, 0\right)=a \gamma^{4}(25 a t / c, 3,4,0) \tag{3.150}
\end{align*}
respectively. The proper time along the trajectory is
{:[tau=int_(0)^(t_(0))sqrt(1-(v(t)//c)^(2))dt=int_(0)^(t_(0))sqrt(1-(5at//c)^(2))dt],[(3.151)=(t_(0))/(2)sqrt(1-(5at_(0)//c)^(2))+(c)/(10 a)arcsin(5at_(0)//c)]:}\begin{align*}
\tau & =\int_{0}^{t_{0}} \sqrt{1-(\boldsymbol{v}(t) / c)^{2}} d t=\int_{0}^{t_{0}} \sqrt{1-(5 a t / c)^{2}} d t \\
& =\frac{t_{0}}{2} \sqrt{1-\left(5 a t_{0} / c\right)^{2}}+\frac{c}{10 a} \arcsin \left(5 a t_{0} / c\right) \tag{3.151}
\end{align*}
1.39
Solution 1: Denote the inertial frame of the observer who measures the length of the rod to be LL by SS, the rest frame of the other observer by S^(')S^{\prime}, and the rest frame of the rod by S^('')S^{\prime \prime}. By the Lorentz contraction formula, the length of the rod in S^('')S^{\prime \prime} is given by
{:(3.152)L^('')=gamma(v)L:}\begin{equation*}
L^{\prime \prime}=\gamma(v) L \tag{3.152}
\end{equation*}
The relative velocity v^(')v^{\prime} between S^(')S^{\prime} and S^('')S^{\prime \prime} is given by relativistic addition of velocities
By the Lorentz contraction formula, the length of the rod in S^(')S^{\prime} is given by
{:(3.154)L^(')=(L^(''))/(gamma(v^(')))=(gamma(v)L)/(gamma(v^(')))=(sqrt(1-v^(2)))/(1+v^(2))*L:}\begin{equation*}
L^{\prime}=\frac{L^{\prime \prime}}{\gamma\left(v^{\prime}\right)}=\frac{\gamma(v) L}{\gamma\left(v^{\prime}\right)}=\frac{\sqrt{1-v^{2}}}{1+v^{2}} \cdot L \tag{3.154}
\end{equation*}
Solution 2: Let x_(A)(t_(A))=(t_(A),vt_(A))x_{A}\left(t_{A}\right)=\left(t_{A}, v t_{A}\right) and x_(B)(t_(B))=(t_(B),vt_(B)+L)x_{B}\left(t_{B}\right)=\left(t_{B}, v t_{B}+L\right) be the worldlines of the ends of the rod. Also, let x_(A)^(')x_{A}^{\prime} and x_(B)^(')x_{B}^{\prime} be events on those worldlines which are simultaneous in S^(')S^{\prime}. Without loss of generality, we can choose x_(A)^(')=(0,0)x_{A}^{\prime}=(0,0). By Lorentz transformation, we find that the worldline x_(B)x_{B} is given by
{:(3.155)x_(B)^(')(t_(B))=gamma(v)(t_(B)(1+v^(2))+Lv,2vt_(B)+L):}\begin{equation*}
x_{B}^{\prime}\left(t_{B}\right)=\gamma(v)\left(t_{B}\left(1+v^{2}\right)+L v, 2 v t_{B}+L\right) \tag{3.155}
\end{equation*}
in the S^(')S^{\prime} frame. Using that x_(A)^(')x_{A}^{\prime} and x_(B)^(')x_{B}^{\prime} are simultaneous in S^(')S^{\prime}, we obtain that x_(B)^(')=x_{B}^{\prime}=x_(B)^(')(tau)x_{B}^{\prime}(\tau) where tau=-Lv//(1+v^(2))\tau=-L v /\left(1+v^{2}\right). Thus, after simplification, the expression for the length of the rod for an observer at rest in S^(')S^{\prime} is given by
{:(3.156)L^(')=x_(B)^(')^(1)-x_(A)^('1)=(sqrt(1-v^(2)))/(1+v^(2))L:}\begin{equation*}
L^{\prime}=x_{B}^{\prime}{ }^{1}-x_{A}^{\prime 1}=\frac{\sqrt{1-v^{2}}}{1+v^{2}} L \tag{3.156}
\end{equation*}
1.40
The 4-velocity of the particle in SS can be written as
By using the relation v^(')=V^(')//V^('0)\boldsymbol{v}^{\prime}=V^{\prime} / V^{\prime 0}, we obtain the velocity of the particle in S^(')S^{\prime} as
Let KK and be the rest frame of the Earth and let K^(')K^{\prime} be the momentary rest frame of the spaceship. Relative to K^(')K^{\prime}, the space ship has the velocity u^(')=0u^{\prime}=0 at the considered time, which, according to transformations of velocities and accelerations, i.e., u^(')=(dx^('))/(dt^('))=(u-v)/(1-(uv)/(c^(2))),quada^(')=(du^('))/(dt^('))=(d^(2)x^('))/(dt^('2))=(a)/(gamma(v)^(3)(1-(uv)/(c^(2)))^(3)),quad gamma(v)-=(1)/(sqrt(1-(v^(2))/(c^(2))))u^{\prime}=\frac{d x^{\prime}}{d t^{\prime}}=\frac{u-v}{1-\frac{u v}{c^{2}}}, \quad a^{\prime}=\frac{d u^{\prime}}{d t^{\prime}}=\frac{d^{2} x^{\prime}}{d t^{\prime 2}}=\frac{a}{\gamma(v)^{3}\left(1-\frac{u v}{c^{2}}\right)^{3}}, \quad \gamma(v) \equiv \frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}},
where u=(dx)/(dt)u=\frac{d x}{d t} and a=(du)/(dt)=(d^(2)x)/(dt^(2))a=\frac{d u}{d t}=\frac{d^{2} x}{d t^{2}}, means that
{:(3.163)u=v quad" and "quad a=a^(')(1-(v^(2))/(c^(2)))^(3//2):}\begin{equation*}
u=v \quad \text { and } \quad a=a^{\prime}\left(1-\frac{v^{2}}{c^{2}}\right)^{3 / 2} \tag{3.163}
\end{equation*}
Since a=(du)/(dt)a=\frac{d u}{d t} and a^(')=ga^{\prime}=g, it follows that
which is a hyperbola in the Minkowski diagram; the motion is said to be hyperbolic.
b) According to the clock hypothesis, it holds that d tau=dt^(')d \tau=d t^{\prime}, where tau\tau is the proper time of the spaceship and t^(')t^{\prime} is the time relative to K^(')K^{\prime}. This means that
If we measure distances in light years and times in years, then c=1c=1 light year/ year and g≃1.05g \simeq 1.05 light years //(" year ")^(2)/(\text { year })^{2}. For x_(A)=2500000x_{A}=2500000 light years, Eq. (3.169) gives the proper time
{:[tau_(A)=(c)/(g)arcosh(1+(gx_(A))/(c^(2)))≃(c)/(g)ln((2gx_(A))/(c^(2)))],[(3.170)~~(1)/(1.05)*ln(2*1.05*2.5*10^(6))" years "~~14.7" years. "]:}\begin{align*}
\tau_{A}=\frac{c}{g} \operatorname{arcosh}\left(1+\frac{g x_{A}}{c^{2}}\right) & \simeq \frac{c}{g} \ln \frac{2 g x_{A}}{c^{2}} \\
& \approx \frac{1}{1.05} \cdot \ln \left(2 \cdot 1.05 \cdot 2.5 \cdot 10^{6}\right) \text { years } \approx 14.7 \text { years. } \tag{3.170}
\end{align*}
Thus, the commander of the spaceship will be almost 55 years old when the spaceship reaches the Andromeda Galaxy.
1.42
With dm < 0d m<0 being the decrease in mass of the rocket and dm^(')d m^{\prime} the mass of the ejecta, the conservation of energy and momentum in the instantaneous rest frame of the rocket takes the form
{:[mc^(2)=(m+dm)gamma(du)c^(2)+dm^(')gamma(-w)c^(2)],[(3.171)≃(m+dm)c^(2)+dm^(')c^(2)gamma(w)],[0=(m+dm)gamma(du)du-wdm^(')gamma(-w)],[(3.172)≃mdu-wdm^(')gamma(w)]:}\begin{align*}
m c^{2} & =(m+d m) \gamma(d u) c^{2}+d m^{\prime} \gamma(-w) c^{2} \\
& \simeq(m+d m) c^{2}+d m^{\prime} c^{2} \gamma(w) \tag{3.171}\\
0 & =(m+d m) \gamma(d u) d u-w d m^{\prime} \gamma(-w) \\
& \simeq m d u-w d m^{\prime} \gamma(w) \tag{3.172}
\end{align*}
where we have kept only terms to linear order in small quantities and dud u is the change in velocity of the rocket in the instantaneous rest frame. Solving for dud u in terms of dmd m, we find that
{:(3.173)du=-(w)/(m)dm:}\begin{equation*}
d u=-\frac{w}{m} d m \tag{3.173}
\end{equation*}
With the relative velocity of the instantaneous rest frame and KK being vv, we find that the change in velocity is given by relativistic addition of velocity
{:[v+dv=(v+du)/(1+(vdu)/(c^(2)))≃(v+du)(1-(vdu)/(c^(2)))≃v+du(1-(v^(2))/(c^(2)))],[(3.174)=v-w(1-(v^(2))/(c^(2)))(dm)/(m)]:}\begin{align*}
v+d v & =\frac{v+d u}{1+\frac{v d u}{c^{2}}} \simeq(v+d u)\left(1-\frac{v d u}{c^{2}}\right) \simeq v+d u\left(1-\frac{v^{2}}{c^{2}}\right) \\
& =v-w\left(1-\frac{v^{2}}{c^{2}}\right) \frac{d m}{m} \tag{3.174}
\end{align*}
where we again keep only linear terms in the small quantities. It follows that
Figure 3.7 Space-station frame (a) and my frame (b).
1.43
In order to obtain the velocity that my friend has in my frame, which will give me the direction, use the formula for relativistic addition of velocities with u_(x)=0u_{x}=0 and u_(y)=vu_{y}=v (which are measured in the space-station frame), i.e.,
where -v-v is also the velocity (along the xx-axis) of the space station in my frame and gamma=(1-v^(2)//c^(2))^(-1//2)\gamma=\left(1-v^{2} / c^{2}\right)^{-1 / 2}. The angle theta\theta is given by (see Figure 3.7)
defined in my frame. Note that in the nonrelativistic limit (i.e., v≪cv \ll c ), the direction is theta_(nr)=-pi//4=-45^(@)\theta_{\mathrm{nr}}=-\pi / 4=-45^{\circ}.
1.44
a) The 4 -velocity of the object is given by U=gamma(u)(c,u)U=\gamma(u)(c, \mathbf{u}) while the 4 -velocity of the frame S^(')S^{\prime} is given by V=gamma(v)(c,-v,0,0)V=\gamma(v)(c,-v, 0,0).
b) In any inertial frame, the 4 -velocity of an object traveling at velocity u\mathbf{u} is given by U=gamma(u)(c,u)U=\gamma(u)(c, \mathbf{u}) as given in a). An object at rest in the inertial frame therefore has 4-velocity V=(c,0)V=(c, \boldsymbol{0}) and as a result
{:(3.180)U*V=c^(2)gamma(u):}\begin{equation*}
U \cdot V=c^{2} \gamma(u) \tag{3.180}
\end{equation*}
As this inner product is a Lorentz invariant, it may be computed in any frame.
c) Taking the inner product of UU and VV as defined in a) we obtain U*V=c^(2)gamma(u^('))=gamma(u)gamma(v)(c^(2)-vu_(1))quad Longrightarrowquad gamma(u^('))=gamma(u)gamma(v)(1-(vu_(1))/(c^(2)))U \cdot V=c^{2} \gamma\left(u^{\prime}\right)=\gamma(u) \gamma(v)\left(c^{2}-v u_{1}\right) \quad \Longrightarrow \quad \gamma\left(u^{\prime}\right)=\gamma(u) \gamma(v)\left(1-\frac{v u_{1}}{c^{2}}\right).
1.45
From the definition of the 4 -velocity, we know that V^(2)=1V^{2}=1. Differentiating this relation with respect to the proper time tau\tau leads to
It follows that the 4 -velocity VV and the 4-acceleration AA are always perpendicular as V*A=0V \cdot A=0.
1.46
a) With the distance to the space station being 1 light day in my rest frame and the station moving toward me with speed c//4c / 4 in the same frame. The distance between the signal and the station will decrease at 5c//45 c / 4. Hence, it will take
{:(3.183)(1" light day ")/(5c//4)=0.8" days, ":}\begin{equation*}
\frac{1 \text { light day }}{5 c / 4}=0.8 \text { days, } \tag{3.183}
\end{equation*}
for the signal to reach the station in my rest frame. The station will then be at a distance of 0.8 light days away from me as this is the distance the signal traveled. The rescue ship then travels toward me with speed 3c//43 c / 4 in the station's rest frame.
Using relativistic addition of velocities, the rescue ship's speed relative to me is
The time taken for the rescue ship to reach me is therefore
{:(3.185)(0.8" light days ")/(16 c//19)=0.95" days. ":}\begin{equation*}
\frac{0.8 \text { light days }}{16 c / 19}=0.95 \text { days. } \tag{3.185}
\end{equation*}
In total, it therefore takes the rescue ship 0.8+0.95=1.750.8+0.95=1.75 days to reach me according to my clock.
1.47
Let us assume that the criminal passes the police officer at t=0t=0 and describe the events from the inertial frame SS. We use a coordinate system such that the worldline of the police ship is given by
as this corresponds to a worldline with proper acceleration of aa (this may be shown by studying the dependence of x_(p)x_{p} on tt for small tt ). As this worldline has x_(p)(t=0)=(1)/(a)x_{p}(t=0)=\frac{1}{a}, the worldline of the criminal must also fulfill x_(c)(t=0)=0x_{c}(t=0)=0 and correspond to the worldline of an object with constant velocity vv. Thus, the criminal's worldline is
The solution t_(1)t_{1} corresponds to the pass where the pursuit starts and t_(2)t_{2} to that where it ends.
a) The criminal is moving with velocity vv relative to SS and is therefore simply time dilated by a factor 1//gamma1 / \gamma. Thus, the pursuit takes
c) By time reversal symmetry in the criminal's rest frame, the relative velocity between the two at the end of the pursuit must be vv (in the opposite direction to when the pursuit started).
For peace of mind, we note that for v rarr0v \rightarrow 0, we find that
a) The relations dt=dX sinh(aT)+Xa cosh(aT)dT,dx=dX cosh(aT)+d t=d X \sinh (a T)+X a \cosh (a T) d T, d x=d X \cosh (a T)+Xa sinh(aT)dT,dy=dY,dz=dZX a \sinh (a T) d T, d y=d Y, d z=d Z give
{:(3.194)ds^(2)=dt^(2)-dx^(2)-dy^(2)-dz^(2)=cdots=(Xa)^(2)dT^(2)-dX^(2)-dY^(2)-dZ^(2):}\begin{equation*}
d s^{2}=d t^{2}-d x^{2}-d y^{2}-d z^{2}=\cdots=(X a)^{2} d T^{2}-d X^{2}-d Y^{2}-d Z^{2} \tag{3.194}
\end{equation*}
i.e., the nonzero components of the metric tensor are g_(TT)=(Xa)^(2)g_{T T}=(X a)^{2} and g_(XX)=g_{X X}=g_(YY)=g_(ZZ)=-1g_{Y Y}=g_{Z Z}=-1.
b) The components of this vector in the astronauts coordinate system are
with (x^('))^(mu^('))=(T,X,Y,Z)\left(x^{\prime}\right)^{\mu^{\prime}}=(T, X, Y, Z) and (x^(v))=(t,x,y,z)\left(x^{v}\right)=(t, x, y, z). We compute the matrix
{:(3.199)([(k^('))^(T)],[(k^('))^(X)],[(k^('))^(Y)],[(k^('))^(Z)])=Lambda([omega],[omega cos(theta)],[0],[omega sin(theta)])=omega([(1//aX)[cosh(aT)-sinh(aT)cos(theta)]],[-sinh(aT)+cosh(aT)cos(theta)],[0],[sin(theta)]):}\left(\begin{array}{c}
\left(k^{\prime}\right)^{T} \tag{3.199}\\
\left(k^{\prime}\right)^{X} \\
\left(k^{\prime}\right)^{Y} \\
\left(k^{\prime}\right)^{Z}
\end{array}\right)=\Lambda\left(\begin{array}{c}
\omega \\
\omega \cos (\theta) \\
0 \\
\omega \sin (\theta)
\end{array}\right)=\omega\left(\begin{array}{c}
(1 / a X)[\cosh (a T)-\sinh (a T) \cos (\theta)] \\
-\sinh (a T)+\cosh (a T) \cos (\theta) \\
0 \\
\sin (\theta)
\end{array}\right)
c) Inserting t(T)=X_(0)sinh(aT),x(T)=X_(0)cosh(aT),y(T)=vTt(T)=X_{0} \sinh (a T), x(T)=X_{0} \cosh (a T), y(T)=v T, and z(T)=0z(T)=0, we obtain
{:(3.200)tau=int_(0)^(T_(0))sqrt(t^(')(T)^(2)-x^(')(T)^(2)-y^(')(T)^(2))dT=T_(0)sqrt((X_(0)a)^(2)-v^(2)):}\begin{equation*}
\tau=\int_{0}^{T_{0}} \sqrt{t^{\prime}(T)^{2}-x^{\prime}(T)^{2}-y^{\prime}(T)^{2}} d T=T_{0} \sqrt{\left(X_{0} a\right)^{2}-v^{2}} \tag{3.200}
\end{equation*}
1.49
By Lorentz invariance of the quantity x^(2)-t^(2)=1//alpha^(2)x^{2}-t^{2}=1 / \alpha^{2}, it follows that x^('2)-t^('2)=x^{\prime 2}-t^{\prime 2}=1//alpha^(2)1 / \alpha^{2}. Differentiating this once with respect to t^(')t^{\prime} gives
for t^(')=0t^{\prime}=0. Thus, the acceleration in S^(')S^{\prime} at time t^(')=0t^{\prime}=0 is the proper acceleration regardless of the relative velocity between the frames SS and S^(')S^{\prime}.
where tau\tau is the proper time since the start of acceleration and the acceleration starts at time t=0t=0. We may of course add an arbitrary constant to the xx-coordinate, but this is irrelevant for our purposes as we can simply put observer AA at x=1//alphax=1 / \alpha. With the light signal being sent from AA at t=t_(0)t=t_{0}, the worldline of the signal is given by
For the case where alphat_(0)≪1\alpha t_{0} \ll 1, we can find a good approximation for the proper time tau\tau by expanding in this parameter with the result
This is no surprise to us as the physical interpretation of the condition is that observer BB has not had enough time to accelerate to a significant velocity, thus the main contribution to the proper time is given by the time t_(0)t_{0}.
In the case alphat_(0)rarr1\alpha t_{0} \rightarrow 1, the time tt as well as the proper time tau\tau for the signal arriving at BB diverge. This is also relatively straightforward to understand as the asymptote of x^(2)-t^(2)=1//alpha^(2)x^{2}-t^{2}=1 / \alpha^{2} is the line x=tx=t. Therefore, if alphat_(0) >= 1\alpha t_{0} \geq 1, then the signal will never reach BB.
Note that the solutions obtained for alphat_(0) > 1\alpha t_{0}>1 are unphysical. They correspond to the intersections of the straight line with the other branch of x^(2)-t^(2)=1//alpha^(2)x^{2}-t^{2}=1 / \alpha^{2}.
1.51
The particles are moving in a circle with constant angular velocity omega\omega. Thus, we can write down an expression for the worldline (in the lab frame) of the particles by using the time tt in the lab frame as the parameter
{:(3.209)x^(mu)=(t","R cos(omega t)","R sin(omega t))^(mu)",":}\begin{equation*}
x^{\mu}=(t, R \cos (\omega t), R \sin (\omega t))^{\mu}, \tag{3.209}
\end{equation*}
where we have assumed the accelerator to have radius RR and suppressed the zz-coordinate which is constant. The velocity of the particles in the lab frame is
since gamma\gamma is a constant. Note that this may also be written in terms of the orbital velocity vv as
{:(3.214)A^(mu)=-gamma^(2)(v^(2))/(R)*(0","cos(vt//R)","sin(vt//R))^(mu):}\begin{equation*}
A^{\mu}=-\gamma^{2} \frac{v^{2}}{R} \cdot(0, \cos (v t / R), \sin (v t / R))^{\mu} \tag{3.214}
\end{equation*}
if using vv as a parameter instead of omega\omega. The proper acceleration aa is given by -a^(2)=A^(2)-a^{2}=A^{2} and thus
{:(3.215)a=sqrt(-A^(2))=gamma^(2)Romega^(2):}\begin{equation*}
a=\sqrt{-A^{2}}=\gamma^{2} R \omega^{2} \tag{3.215}
\end{equation*}
The eigentime required for the particles to complete one orbit is simply given by the relation between the eigentime and the lab time
where T=2pi//omegaT=2 \pi / \omega is the time to complete the orbit as measured in the lab, by integrating the differential relation taking into account the fact that gamma\gamma is constant.
where P=MVP=M V is the 4 -momentum and tau\tau the proper time of the object. The internal energy can be found by squaring the 4-momentum P^(2)=M^(2)V^(2)=M^(2)P^{2}=M^{2} V^{2}=M^{2}, since the square of the 4 -velocity is equal to one. It follows that
{:(3.218)(dP^(2))/(d tau)=2P*(dP)/(d tau)=2M(dM)/(d tau):}\begin{equation*}
\frac{d P^{2}}{d \tau}=2 P \cdot \frac{d P}{d \tau}=2 M \frac{d M}{d \tau} \tag{3.218}
\end{equation*}
where dP//d tau=Fd P / d \tau=F is the 4-force and dM//d taud M / d \tau is the sought derivative of the internal energy with respect to the proper time. Using P=MVP=M V, we can solve for this quantity as
where gamma=V*U\gamma=V \cdot U is the gamma factor for the relative velocity vv between objects moving with the 4 -velocities UU and VV. (Alternatively, we note that there is a frame where U=(1,0)U=(1,0). In this frame, V=gamma(1,v)V=\gamma(1, \boldsymbol{v}), where v\boldsymbol{v} is the velocity of the object in this frame.)
By using the product rule on the derivative F=dP//d tauF=d P / d \tau, we find that
where A=dV//d tauA=d V / d \tau is the 4-acceleration of the object. The relation between the 4-acceleration AA and the proper acceleration alpha\alpha is
{:(3.222)alpha=(f)/(M)gamma v:}\begin{equation*}
\alpha=\frac{f}{M} \gamma v \tag{3.222}
\end{equation*}
In particular, we note that alpha=0\alpha=0 if v=0v=0, which is stating that the object is not accelerating if the 4 -force is parallel to the 4 -velocity, as expected.
1.53
With the 4-force being the derivative of the 4-momentum with respect to the proper time, we find that
where V^(mu)V^{\mu} is the 4 -velocity and A^(mu)A^{\mu} the 4 -acceleration. We can find the rate of change in the rest energy by taking the inner product with the 4 -velocity, resulting in
{:(3.224)F*V=(dm)/(d tau)V^(2)+mA*V=(dm)/(d tau):}\begin{equation*}
F \cdot V=\frac{d m}{d \tau} V^{2}+m A \cdot V=\frac{d m}{d \tau} \tag{3.224}
\end{equation*}
since V^(2)=1V^{2}=1 and A*V=0A \cdot V=0. Using the given expression for F^(mu)F^{\mu} and that V^(mu)=gamma(1,v)^(mu)V^{\mu}=\gamma(1, \boldsymbol{v})^{\mu}, we now obtain
The force is a pure force only if the rate dm//d tau=0d m / d \tau=0, corresponding to f*v=0\boldsymbol{f} \cdot \boldsymbol{v}=0. Therefore, the force is a pure force if ff and vv are orthogonal.
We now use that the 4 -velocity in S^(')S^{\prime} is given by V=gamma(1,-v,0)V=\gamma(1,-v, 0) to identify this with the general expression for the 4 -acceleration
For a fixed proper acceleration alpha\alpha, the acceleration in the instantaneous rest frame is given by a_(0x)=alpha cos(theta)a_{0 x}=\alpha \cos (\theta) and a_(0y)=alpha sin(theta)a_{0 y}=\alpha \sin (\theta), where theta\theta is the angle between the acceleration and the xx-direction. This results in an acceleration
Since gamma >= 1\gamma \geq 1, the expression in the parenthesis varies between one [when sin^(2)(theta)=1\sin ^{2}(\theta)=1 ] and gamma^(-2)\gamma^{-2} [when {:cos^(2)(theta)=1]\left.\cos ^{2}(\theta)=1\right]. Consequently, we find that
which is the 4 -velocity of an object at rest in S^(')S^{\prime}. Similarly, the x^(')x^{\prime}-component of the 3 -momentum can be obtained by taking the inner product with the 4 -vector
To see this, we note that the 4 -momentum in S^(')S^{\prime} is given by P^(')=(E^('),p^('))P^{\prime}=\left(E^{\prime}, \boldsymbol{p}^{\prime}\right), resulting in
Furthermore, the velocity of a particle in an arbitrary frame is given by v=p//Ev=p / E, where p\boldsymbol{p} is the 3-momentum and EE the energy. In particular, in S^(')S^{\prime}, this results in the velocity component in the x^(')x^{\prime}-direction being
From the expression for UU in S^(')S^{\prime}, we therefore have gamma(u^('))^(2)=([1+vu cos(theta)]^(2))/((1-v^(2))(1-u^(2)))quad Longrightarrowquadu^(')=(sqrt([1+vu cos(theta)]^(2)-(1-v^(2))(1-u^(2))))/(1+vu cos(theta))\gamma\left(u^{\prime}\right)^{2}=\frac{[1+v u \cos (\theta)]^{2}}{\left(1-v^{2}\right)\left(1-u^{2}\right)} \quad \Longrightarrow \quad u^{\prime}=\frac{\sqrt{[1+v u \cos (\theta)]^{2}-\left(1-v^{2}\right)\left(1-u^{2}\right)}}{1+v u \cos (\theta)}.
The tangent of the angle theta^(')\theta^{\prime}, can be expressed as the ratio between U^(2^('))U^{2^{\prime}} and U^(1^('))U^{1^{\prime}}, i.e.,
In the nonrelativistic limit u,v≪1u, v \ll 1, the above expressions become
{:(3.254)u^(')=sqrt(v^(2)+u^(2)+2vu cos(theta))quad" and "quad tan(theta^('))=(u sin(theta))/(u cos(theta)+v):}\begin{equation*}
u^{\prime}=\sqrt{v^{2}+u^{2}+2 v u \cos (\theta)} \quad \text { and } \quad \tan \left(\theta^{\prime}\right)=\frac{u \sin (\theta)}{u \cos (\theta)+v} \tag{3.254}
\end{equation*}
to leading order. The relation for the speed u^(')u^{\prime} is just the cosine theorem for classical addition of velocities and the expression for tan(theta^('))\tan \left(\theta^{\prime}\right) is the classical aberration formula.
and its velocity by v(t)=(dx)/(dt)=atv(t)=\frac{d x}{d t}=a t. The speed v_(0)v_{0} is obtained at time t_(0)=(v_(0))/(a)t_{0}=\frac{v_{0}}{a}. The proper time to reach this speed is therefore
{:(3.257)tau(v_(0))=int_(0)^(t_(0))sqrt(1-v(t)^(2))dt=int_(0)^(t_(0))sqrt(1-a^(2)t^(2))dt:}\begin{equation*}
\tau\left(v_{0}\right)=\int_{0}^{t_{0}} \sqrt{1-v(t)^{2}} d t=\int_{0}^{t_{0}} \sqrt{1-a^{2} t^{2}} d t \tag{3.257}
\end{equation*}
Substituting at=sin thetaa t=\sin \theta, we find that dt=(1)/(a)cos theta d thetad t=\frac{1}{a} \cos \theta d \theta, and therefore,
where theta_(0)=arcsin(at_(0))=arcsin v_(0)\theta_{0}=\arcsin \left(a t_{0}\right)=\arcsin v_{0}. Inserting the expression for theta_(0)\theta_{0} leads to
The speed of a light signal relative to the water is given by u_(0)=c//nu_{0}=c / n, where nn is the refractive index of water. Meanwhile, the speed of the water relative to the lab frame is vv and application of the formula for relativistic addition of velocities then results in the speed of the light signal relative to the lab frame
{:(3.266)u=(u_(0)+v)/(1+(u_(0)v)/(c^(2)))≃u_(0)+kv:}\begin{equation*}
u=\frac{u_{0}+v}{1+\frac{u_{0} v}{c^{2}}} \simeq u_{0}+k v \tag{3.266}
\end{equation*}
where k=1-(u_(0)^(2))/(c^(2))=1-(1)/(n^(2))k=1-\frac{u_{0}^{2}}{c^{2}}=1-\frac{1}{n^{2}} and only linear terms in v≪cv \ll c have been kept, which is Fizeau's result.
1.60
See the solution to Problem 1.59, i.e., Fizeau's result. In the case of the water moving perpendicular to the light, we assume that the water is moving in the xx-direction with speed vv and the light is moving in the yy-direction with speed c//nc / n (when the water is standing still). In this case, the formula for addition of velocities reads
In other words, the correction enters only at second order in the water speed.
1.61
The redshift is maximal for when the source moves directly away from the observer. A lower bound for the speed of 3 C 9 is therefore determined as
{:(3.268)u_(min)=(x^(2)-1)/(x^(2)+1)c≃0.80 c","quad" where "x=(3600)/(1215):}\begin{equation*}
u_{\min }=\frac{x^{2}-1}{x^{2}+1} c \simeq 0.80 c, \quad \text { where } x=\frac{3600}{1215} \tag{3.268}
\end{equation*}
is the ratio of the observed and emitted wavelengths, by solving for xx from the relativistic Doppler shift formula.
1.62
The electromagnetic wave is E(x)=E_(0)sin 2pi((x^(1))/(lambda)-vt)=sin((2pix^(1))/(lambda)-2pi nu t)E(x)=E_{0} \sin 2 \pi\left(\frac{x^{1}}{\lambda}-v t\right)=\sin \left(\frac{2 \pi x^{1}}{\lambda}-2 \pi \nu t\right). The argument can be rewritten as follows
{:[(2pix^(1))/(lambda)-2pi vt=-2pi vt+(2pix^(1))/(lambda)=-(2pi vt-(2pix^(1))/(lambda))=-(2pi vt-(2pi v)/(lambda v)x^(1))],[={omega=2pi v" and "c=lambda nu}=-(omega t-(omega )/(c)x^(1))=-((omega )/(c)ct-(omega )/(c)x^(1))],[={x^(0)=ct}=-((omega )/(c)x^(0)-(omega )/(c)x^(1))],[(3.269)=-((omega )/(c),(omega )/(c),0,0)*(x^(0),x^(1),x^(2),x^(3))=-k_(mu)x^(mu)]:}\begin{align*}
\frac{2 \pi x^{1}}{\lambda}-2 \pi v t & =-2 \pi v t+\frac{2 \pi x^{1}}{\lambda}=-\left(2 \pi v t-\frac{2 \pi x^{1}}{\lambda}\right)=-\left(2 \pi v t-\frac{2 \pi v}{\lambda v} x^{1}\right) \\
& =\{\omega=2 \pi v \text { and } c=\lambda \nu\}=-\left(\omega t-\frac{\omega}{c} x^{1}\right)=-\left(\frac{\omega}{c} c t-\frac{\omega}{c} x^{1}\right) \\
& =\left\{x^{0}=c t\right\}=-\left(\frac{\omega}{c} x^{0}-\frac{\omega}{c} x^{1}\right) \\
& =-\left(\frac{\omega}{c}, \frac{\omega}{c}, 0,0\right) \cdot\left(x^{0}, x^{1}, x^{2}, x^{3}\right)=-k_{\mu} x^{\mu} \tag{3.269}
\end{align*}
where k=((omega )/(c),(omega )/(c),0,0)k=\left(\frac{\omega}{c}, \frac{\omega}{c}, 0,0\right) and x=(x^(0),x^(1),x^(2),x^(3))x=\left(x^{0}, x^{1}, x^{2}, x^{3}\right). Thus, we can write E(x)=E(x)=sin(-k_(mu)x^(mu))=-sin k_(mu)x^(mu)\sin \left(-k_{\mu} x^{\mu}\right)=-\sin k_{\mu} x^{\mu}.
The wave vector kk is lightlike, since k^(2)=k_(mu)k^(mu)=((omega )/(c),(omega )/(c),0,0)*((omega )/(c),(omega )/(c),0,0)=k^{2}=k_{\mu} k^{\mu}=\left(\frac{\omega}{c}, \frac{\omega}{c}, 0,0\right) \cdot\left(\frac{\omega}{c}, \frac{\omega}{c}, 0,0\right)=((omega )/(c))^(2)-((omega )/(c))^(2)-0^(2)-0^(2)=0\left(\frac{\omega}{c}\right)^{2}-\left(\frac{\omega}{c}\right)^{2}-0^{2}-0^{2}=0.
In K^(')K^{\prime}, we have E^(')(x^('))=-E_(0)sin k_(mu)^(')x^('mu)=-E_(0)sin k_(mu)x^(mu)=E(x)E^{\prime}\left(x^{\prime}\right)=-E_{0} \sin k_{\mu}^{\prime} x^{\prime \mu}=-E_{0} \sin k_{\mu} x^{\mu}=E(x). Since this is Lorentz invariant, we find that k^(')=Lambda kk^{\prime}=\Lambda k, where Lambda\Lambda is a Lorentz transformation, and (omega^('))/(c)=k^('0)=k^(0)cosh theta-k^(1)sinh theta=(omega )/(c)cosh theta-(omega )/(c)sinh theta=(omega )/(c)(cosh theta-sinh theta)\frac{\omega^{\prime}}{c}=k^{\prime 0}=k^{0} \cosh \theta-k^{1} \sinh \theta=\frac{\omega}{c} \cosh \theta-\frac{\omega}{c} \sinh \theta=\frac{\omega}{c}(\cosh \theta-\sinh \theta).
Using the definitions of the hyperbolic functions and the fact that omega=2pi nu\omega=2 \pi \nu and omega^(')=2pinu^(')\omega^{\prime}=2 \pi \nu^{\prime}, we obtain the answer
which is the formula for the Doppler shift. If we instead use the relations cosh theta=\cosh \theta=gamma(v)\gamma(v) and sinh theta=(v)/(c)gamma(v)\sinh \theta=\frac{v}{c} \gamma(v), where gamma(v)-=(1)/(sqrt(1-((v)/(c))^(2)))\gamma(v) \equiv \frac{1}{\sqrt{1-\left(\frac{v}{c}\right)^{2}}}, we instead obtain
From the statement that the GRB has twice the duration as a nearby GRB, we conclude that the GRB is time dilated due to its motion and that gamma=2\gamma=2. From this relation, we can solve for the velocity of the GRB and obtain
where we assume, without loss of generality, that the objects move parallel to the xx-axis and that the first object is at the origin at time t=0;y_(j)(t)=z_(j)(t)=0t=0 ; y_{j}(t)=z_{j}(t)=0, for j=1,2j=1,2, and L > 0L>0 is the distance between the objects at time t=0t=0, of course. The frame of reference K^(')K^{\prime} of the observer on the first object is related to KK by a Lorentz transformation
and y^(')=y,z^(')=zy^{\prime}=y, z^{\prime}=z with v=c//2v=c / 2 so that x_(1)^(')(t)=0x_{1}^{\prime}(t)=0.
a) We parametrize the trajectory of the second object in KK as
to the photon. The wavelength is shorter: it has been shifted from red to UV, i.e., the frequency of the photon is higher.
1.65
a) In the rest frame of the observer, we can choose coordinates such that the light source is on the positive xx-axis. The 4 -velocity of the light source can then be written as V=gamma(v)(1,v cos(theta),v sin(theta),0)V=\gamma(v)(1, v \cos (\theta), v \sin (\theta), 0) and the 4-frequency of the light pulse is given by N=omega(1,-1,0,0)N=\omega(1,-1,0,0). The frequency in the rest frame of the source is then obtained as
In particular, when the light source is moving directly away from the observer, we have cos(theta)=1\cos (\theta)=1 and therefore recover the relativistic Doppler formula
Thus, for small velocities, the angle theta\theta is close to 90^(@)90^{\circ}, but the source must move slightly toward the observer, whereas for velocities close to the speed of light, the source must move almost straight toward the observer.
1.66
We have a rotating disk with two observers O_(1)O_{1} and O_(2)O_{2} at different radii r_(1)r_{1} and r_(2)r_{2}, respectively. Setting c=1c=1 and omitting the trivial zz-direction, introducing polar coordinates on the inertial frame leads to
{:(3.295)ds^(2)=dt^(2)-dr^(2)-r^(2)dphi^(2):}\begin{equation*}
d s^{2}=d t^{2}-d r^{2}-r^{2} d \phi^{2} \tag{3.295}
\end{equation*}
We now introduce a rotating frame with phi_(r)=phi+Omega t\phi_{r}=\phi+\Omega t, which leads to phi=phi_(r)-Omega t\phi=\phi_{r}-\Omega t and find that
{:[ds^(2)=dt^(2)-dr^(2)-r^(2)dphi^(2)=dt^(2)-dr^(2)-r^(2)(dphi_(r)-Omega dt)^(2)],[(3.296)=(1-r^(2)Omega^(2))dt^(2)+r^(2)Omega dtdphi_(r)-dr^(2)-r^(2)dphi_(r)^(2)]:}\begin{align*}
d s^{2} & =d t^{2}-d r^{2}-r^{2} d \phi^{2}=d t^{2}-d r^{2}-r^{2}\left(d \phi_{r}-\Omega d t\right)^{2} \\
& =\left(1-r^{2} \Omega^{2}\right) d t^{2}+r^{2} \Omega d t d \phi_{r}-d r^{2}-r^{2} d \phi_{r}^{2} \tag{3.296}
\end{align*}
Thus, we observe that we have a nontrivial time-time component of the metric, i.e.,
where Delta t\Delta t is the time difference between the emission of the wave fronts in the inertial frame. Therefore, using that the observed angular frequency omega\omega is inversely proportional to the time period, we obtain
where the coordinate system has been arranged such that the third spatial component is zero (which is why it has been omitted). Here, vv is the frequency of the wave in the rest frame of the medium. In the rest frame of the source S^(')S^{\prime}, the frequency is given by the zeroth component of the 4 -frequency in that frame
where VV is the 4 -velocity of the source itself. This expression is a Lorentz scalar and may be computed in any frame. In particular, in SS, the 4 -velocity of the source is
where we have introduced c_(i)=cos theta_("in "),s_(i)=sin theta_("in "),c_(o)=cos theta_("out "),s_(o)=sin theta_("out ")c_{i}=\cos \theta_{\text {in }}, s_{i}=\sin \theta_{\text {in }}, c_{o}=\cos \theta_{\text {out }}, s_{o}=\sin \theta_{\text {out }}, and used a coordinate system such that the mirror is moving in the xx-direction and the light is not propagating in the zz-direction, which we therefore have omitted. In the rest frame of the mirror S^(')S^{\prime}, the incident angle is equal to the reflected angle. Furthermore, the frequencies of the incoming and reflected light are the same, which means that
where c^(')=cos theta^('),s^(')=sin theta^(')c^{\prime}=\cos \theta^{\prime}, s^{\prime}=\sin \theta^{\prime}. Since pp and kk are related to p^(')p^{\prime} and k^(')k^{\prime} by Lorentz transformation, we also have
For v=-c_(i)=-cos theta_(in)v=-c_{i}=-\cos \theta_{\mathrm{in}}, the mirror is moving away from the light at the same speed that the light is approaching the mirror. As a result, the light never reaches the mirror. In S^(')S^{\prime}, the light is moving parallel to the mirror. The reflection angle in SS approaches c_(o)=-c_(i)c_{o}=-c_{i}, which simply means that the light continues in a straight line. For values v <= c_(i)v \leq c_{i}, the mirror outruns the light and there is no reflection.
1.69
In the rest frame S^(')S^{\prime} of the medium, the 4 -frequencies of the light waves are given by
where c^(')=cos theta^('),s^(')=sin theta^('),s^('')=sin theta^(''),c^('')=cos theta^('')c^{\prime}=\cos \theta^{\prime}, s^{\prime}=\sin \theta^{\prime}, s^{\prime \prime}=\sin \theta^{\prime \prime}, c^{\prime \prime}=\cos \theta^{\prime \prime} and the relation between theta^(')\theta^{\prime} and theta^('')\theta^{\prime \prime} is given by Snell's law sin theta^(')=n sin theta^('')\sin \theta^{\prime}=n \sin \theta^{\prime \prime}. Here NN is the 4 -frequency of the wave before entering the medium and N\mathcal{N} is the 4 -frequency after entering the medium. Expressing theta^('')\theta^{\prime \prime} in terms of theta^(')\theta^{\prime}, we obtain
which has the solutions n^(2)=1n^{2}=1 and n^(2)=1+2c^(')vgamma^(2)n^{2}=1+2 c^{\prime} v \gamma^{2}. Since n > 1n>1 was given, the latter of these is the sought solution and we obtain
{:(3.320)n=sqrt(1+2c^(')vgamma^(2)):}\begin{equation*}
n=\sqrt{1+2 c^{\prime} v \gamma^{2}} \tag{3.320}
\end{equation*}
Note that the solution n=1n=1 corresponds to the scenario where the medium has the same refractive index as vacuum, i.e., is vacuum. In this case, it does not matter how fast the medium moves and the light wave will continue undisturbed.
1.70
We can find the angular frequency omega_(0)\omega_{0} of the wave in the source rest frame by multiplying the 4 -frequency NN by the 4 -velocity of the source. This results in
When v rarr uv \rightarrow u, the source is moving essentially at the same velocity as the wave speed in the medium and in the same direction as the wave. As a result, the wave train is compressed and the frequency approaches infinity. This is the same behavior as obtained for the classical Doppler shift. When v rarr-uv \rightarrow-u, the source moves in the opposite direction to the wave, resulting in a frequency omega rarromega_(0)//(2gamma)\omega \rightarrow \omega_{0} /(2 \gamma). The gamma\gamma factor appears due to the time dilation of the source, while the factor of two results from the wavelength being doubled due to the motion of the source. The only difference here to the classical Doppler shift is the appearance of the factor gamma\gamma, describing the time dilation due to the motion of the source.
1.71
From the form of the worldline for observer BB, we find that
{:(3.323)xx^(˙)-tt^(˙)=0:}\begin{equation*}
x \dot{x}-t \dot{t}=0 \tag{3.323}
\end{equation*}
by differentiating with respect to the proper time. It directly follows that v=v=x^(˙)//t^(˙)=t//x\dot{x} / \dot{t}=t / x. Solving for xx in terms of tt and inserting it into this expression now gives
The 4 -frequency of the signal is given by Omega=omega(1,1)\Omega=\omega(1,1) in SS and the frequency observed by BB can be found by taking the inner product of this 4 -frequency with the 4 -velocity of BB. We find that
The rest energy of an electron is E_(0)=m_(0)c^(2)≃0.51MeVE_{0}=m_{0} c^{2} \simeq 0.51 \mathrm{MeV}, where m_(0)m_{0} is the rest mass of the electron. Thus, the total energy after acceleration is therefore given by
{:(3.331)v=csqrt(1-((E_(0))/(E))^(2))=csqrt(1-((0.51)/(1.51))^(2))≃0.94 c:}\begin{equation*}
v=c \sqrt{1-\left(\frac{E_{0}}{E}\right)^{2}}=c \sqrt{1-\left(\frac{0.51}{1.51}\right)^{2}} \simeq 0.94 c \tag{3.331}
\end{equation*}
i.e., the final velocity of the electron is v≃0.94 cv \simeq 0.94 c.
1.73
a) If the equation p_(e^(-))+p_(e^(+))=k_(gamma)p_{e^{-}}+p_{e^{+}}=k_{\gamma} i.e., conservation of 4-momentum (where p_(e^(-))p_{e^{-}}is the 4 -momentum of the electron, p_(e^(+))p_{e^{+}}is the 4 -momentum of the positron, and k_(gamma)k_{\gamma} is the 4 -momentum of the photon), is squared and the left-hand side is calculated in the rest frame of the electron, then the relation
is obtained, where m_(e)m_{e} is the rest mass of the electron (or positron) and E_(e^(+))E_{e^{+}}is the total energy of the positron relative to the rest frame of the electron. This proves that this process is incompatible with conservation of 4 -momentum (i.e., conservation of energy and momentum) as all of the quantities on the left-hand side are strictly positive.
Alternatively, this can also be seen in an inertial frame where the spatial parts of the total momenta (i.e., the 3-momenta) of the electron and the positron are zero before the collision. In this frame, the 4 -momenta of the electron and the positron before the collision are
{:(3.333)p_(e^(-))=(sqrt(m_(e)^(2)+p^(2)),p)quad" and "quadp_(e^(+))=(sqrt(m_(e)^(2)+p^(2)),-p):}\begin{equation*}
p_{e^{-}}=\left(\sqrt{m_{e}^{2}+\boldsymbol{p}^{2}}, \boldsymbol{p}\right) \quad \text { and } \quad p_{e^{+}}=\left(\sqrt{m_{e}^{2}+p^{2}},-\boldsymbol{p}\right) \tag{3.333}
\end{equation*}
where p=(p_(1),p_(2),p_(3))\boldsymbol{p}=\left(p_{1}, p_{2}, p_{3}\right). Let k_(gamma)=(|k|,k)k_{\gamma}=(|\boldsymbol{k}|, \boldsymbol{k}) be the lightlike 4-momentum of the photon. Then, conservation of 4-momentum p_(e^(-))+p_(e^(+))=k_(gamma)p_{e^{-}}+p_{e^{+}}=k_{\gamma} implies that
{:(3.334)2sqrt(m_(e)^(2)+p^(2))=|k|quad" and "quad0=k:}\begin{equation*}
2 \sqrt{m_{e}^{2}+p^{2}}=|k| \quad \text { and } \quad 0=k \tag{3.334}
\end{equation*}
These two conditions clearly contradict each other.
b) Now, let p_("in ")p_{\text {in }} and p_("out ")p_{\text {out }} be the 4 -momentum of an electron before and after emitting a photon, respectively. In addition, let the photon have 4-momentum kk. For this process to conserve the total 4-momentum, the relation
where omega\omega is the photon energy in the rest frame of the electron after emitting the photon. This cannot hold for any nonzero photon energy omega\omega.
c) In this case, we have p_(mu)+p_(mu)^(')=k_(mu)+k_(mu)^(')p_{\mu}+p_{\mu}^{\prime}=k_{\mu}+k_{\mu}^{\prime}, where (k_(mu))=(|k|,k)\left(k_{\mu}\right)=(|\boldsymbol{k}|, \boldsymbol{k}) and (k_(mu)^('))=\left(k_{\mu}^{\prime}\right)=(|k^(')|,k^('))\left(\left|\boldsymbol{k}^{\prime}\right|, \boldsymbol{k}^{\prime}\right) are the 4-momenta of the two photons, respectively. This implies that
{:(3.337)2sqrt(m_(e)^(2)+p^(2))=|k|+|k^(')|quad" and "quad0=k+k^('):}\begin{equation*}
2 \sqrt{m_{e}^{2}+p^{2}}=|\boldsymbol{k}|+\left|\boldsymbol{k}^{\prime}\right| \quad \text { and } \quad 0=\boldsymbol{k}+\boldsymbol{k}^{\prime} \tag{3.337}
\end{equation*}
which clearly has a nontrivial solution k^(')=-kk^{\prime}=-\boldsymbol{k} and |k|=sqrt(m_(e)^(2)+p^(2))|\boldsymbol{k}|=\sqrt{m_{e}^{2}+p^{2}}. Thus, the answer to the question is "yes."
1.74
It holds that M^(2)=P_("before ")^(2)=P_("after ")^(2)=(p_(a)+p_(b))^(2)M^{2}=P_{\text {before }}^{2}=P_{\text {after }}^{2}=\left(p_{a}+p_{b}\right)^{2}, where p_(a)=(E_(a),p_(a))p_{a}=\left(E_{a}, \boldsymbol{p}_{a}\right) and p_(b)=p_{b}=(m_(b),0)\left(m_{b}, 0\right) in the rest frame of bb. Solving this equation for |p_(a)|\left|p_{a}\right|, using E_(a)=sqrt(m_(a)^(2)+p_(a)^(2))E_{a}=\sqrt{m_{a}^{2}+p_{a}^{2}}, gives
Conservation of 4 -momentum tells us that P=p_(1)+p_(2)P=p_{1}+p_{2}, where PP is the 4 -momentum of the new particle and p_(1)p_{1} and p_(2)p_{2} that of the initial two particles.
Squaring this relation we find that
In the rest frame of particle 2, we have p_(1)=m gamma(1,v_(1))p_{1}=m \gamma\left(1, \boldsymbol{v}_{1}\right) and p_(2)=(m,0)p_{2}=(m, \mathbf{0}), leading to
In addition, using that v=p//E\boldsymbol{v}=\boldsymbol{p} / E for any particle and that the final 3-momentum is equal to the 3 -momentum of particle 1 in the rest frame of particle 2, we obtain
where gamma(u_(i))-=1//sqrt(1-u_(i)^(2)),i=1,2\gamma\left(u_{i}\right) \equiv 1 / \sqrt{1-u_{i}^{2}}, i=1,2. The energies of the two particles can then be rewritten as
Assuming that the two particles move along the xx-axis in the frame of the observer and using the expressions for the 3-momenta of the two particles, we find that
which is the required expression in terms of the relative velocity vv.
c) In situation 1(u_(1)=0)1\left(u_{1}=0\right) we have u_(2)=vu_{2}=v. The total energy is therefore
In situation 2(m_(1)gamma(u_(1))u_(1)=-m_(2)gamma(u_(2))u_(2))2\left(m_{1} \gamma\left(u_{1}\right) u_{1}=-m_{2} \gamma\left(u_{2}\right) u_{2}\right), our frame is instead the rest frame of the new particle and the total energy is therefore
where gamma=gamma(v)\gamma=\gamma(v), with equality only if v=0v=0. Therefore, more energy will be required in the rest frame of one of the particles than in the center-of-mass frame. For v≪1v \ll 1, keeping only terms up to second order in vv, we find that
{:(3.369)Delta E≃(m_(2)-mu)(v^(2))/(2)=(m_(2)^(2)v^(2))/(2(m_(1)+m_(2))):}\begin{equation*}
\Delta E \simeq\left(m_{2}-\mu\right) \frac{v^{2}}{2}=\frac{m_{2}^{2} v^{2}}{2\left(m_{1}+m_{2}\right)} \tag{3.369}
\end{equation*}
where mu\mu is the reduced mass mu=m_(1)m_(2)//(m_(1)+m_(2))\mu=m_{1} m_{2} /\left(m_{1}+m_{2}\right) of the two-particle system.
1.78
Conservation of 4-momentum gives p_(pi)=p_(mu)+p_(nu)p_{\pi}=p_{\mu}+p_{\nu}, where p_(pi)=(E_(pi),p,0,0)p_{\pi}=\left(E_{\pi}, p, 0,0\right) and p_(mu)=(E_(mu),0,( hat(p)),0)p_{\mu}=\left(E_{\mu}, 0, \hat{p}, 0\right). Here (p,0,0)(p, 0,0) and (0, hat(p),0)(0, \hat{p}, 0) are the 3-momenta of the pion (in the xx-direction) and the muon (in the yy-direction), respectively, which are, however, not important for this problem. Taking the square of the 4-momentum relation after moving p_(mu)p_{\mu} to the left-hand side, we find that
where gamma(v)-=1//sqrt(1-v^(2))\gamma(v) \equiv 1 / \sqrt{1-v^{2}} and vv is the velocity of the incoming pion.
1.79
Let the 4-momenta of the pion, the electron, and the antineutrino be p_(pi),p_(e)p_{\pi}, p_{e}, and p_(v)p_{v}, respectively. We find from energy-momentum conservation the relation
In the rest frame of the electron, we have p_(e)=(m,0)p_{e}=(m, \mathbf{0}) and p_(v)=(E_(v),p)p_{v}=\left(E_{v}, \mathbf{p}\right), where mm is the mass of the electron and E_(v)=sqrt(m_(v)^(2)+p^(2))E_{v}=\sqrt{m_{v}^{2}+\mathbf{p}^{2}} and p\mathbf{p} are the total energy and the
3-momentum of the antineutrino, respectively, m_(v)m_{v} being the mass of the antineutrino. Thus, we obtain
{:(3.373)m_(pi)^(2)=(p_(e)+p_(v))^(2)=p_(e)^(2)+p_(v)^(2)+2p_(e)*p_(v)=m^(2)+m_(v)^(2)+2mE_(v):}\begin{equation*}
m_{\pi}^{2}=\left(p_{e}+p_{v}\right)^{2}=p_{e}^{2}+p_{v}^{2}+2 p_{e} \cdot p_{v}=m^{2}+m_{v}^{2}+2 m E_{v} \tag{3.373}
\end{equation*}
and hence, we have E_(v)=Delta//(2m)E_{v}=\Delta /(2 m), where Delta-=m_(pi)^(2)-m^(2)-m_(v)^(2)\Delta \equiv m_{\pi}^{2}-m^{2}-m_{v}^{2}. Using that the absolute value of the 3 -momentum of the antineutrino is
For the limiting value of vv as the rest mass of the antineutrino goes to zero, since lim_(m_(v)rarr0)Delta=m_(pi)^(2)-m^(2)\lim _{m_{v} \rightarrow 0} \Delta=m_{\pi}^{2}-m^{2}, we find that v rarr1v \rightarrow 1 as m_(v)rarr0m_{v} \rightarrow 0.
1.80
a) Consider the reaction pi^(+)longrightarrowmu^(+)+v_(mu)\pi^{+} \longrightarrow \mu^{+}+v_{\mu}. Conservation of 4-momentum gives that
Since the pion decays at rest, the absolute value of the neutrino momentum must equal that of the muon momentum. The energy-momentum relation for the pion therefore yields
after inserting the expression for E_(mu)E_{\mu} and solving for p_(nu)p_{\nu}.
b) In the rest frame of the pion, the muon will travel the distance s-=gamma(v_(mu))v_(mu)tau_(mu)=s \equiv \gamma\left(v_{\mu}\right) v_{\mu} \tau_{\mu}=p_(mu)tau_(mu)//m_(mu)p_{\mu} \tau_{\mu} / m_{\mu} before it decays. We therefore find that
since the muon and neutrino momenta are equal in magnitude.
1.81
In the rest frame of the decaying pion, the solution to Problem 1.80 resulted in
{:(3.383)E_(mu)=(m_(pi)^(2)+m_(mu)^(2))/(2m_(pi))quad" and "quad p=(m_(pi)^(2)-m_(mu)^(2))/(2m_(pi)):}\begin{equation*}
E_{\mu}=\frac{m_{\pi}^{2}+m_{\mu}^{2}}{2 m_{\pi}} \quad \text { and } \quad p=\frac{m_{\pi}^{2}-m_{\mu}^{2}}{2 m_{\pi}} \tag{3.383}
\end{equation*}
for the muon energy and momentum, respectively. It follows that the 4-momentum of the muon in the rest frame of the pion is given by p_(mu)=(E_(mu),p)p_{\mu}=\left(E_{\mu}, p\right). However, we want to compute the energy of the muon in the rest frame of the Earth, and thus, we must Lorentz transform p_(mu)p_{\mu} to this frame. The Lorentz transformation is in the opposite direction of the motion of the muon in the rest frame of the pion with velocity v=sqrt(E_(pi)^(2)-m_(pi)^(2))//E_(pi)v=\sqrt{E_{\pi}^{2}-m_{\pi}^{2}} / E_{\pi}, where E_(pi)E_{\pi} is the energy of the pion in the rest frame of the Earth. It follows that in the rest frame of the Earth, we have
We study the decay R longrightarrowmu^(+)+mu^(-)R \longrightarrow \mu^{+}+\mu^{-}. In this decay, the total 4 -momentum must be preserved, and thus, we have
If we place our coordinate system in such a way that the mu^(+)\mu^{+}is traveling in the xx-direction and the mu^(-)\mu^{-}in the yy-direction (this is possible, since the angle between the directions is 90^(@)90^{\circ} ), then the 4 -momentum of the muons will be
{:(3.388)p_(mu^(+))=(E","p","0","0)quad" and "quadp_(mu^(-))=(E","0","p","0):}\begin{equation*}
p_{\mu^{+}}=(E, p, 0,0) \quad \text { and } \quad p_{\mu^{-}}=(E, 0, p, 0) \tag{3.388}
\end{equation*}
respectively, where p=2.2GeVp=2.2 \mathrm{GeV} and E=sqrt(p^(2)+m_(mu)^(2))E=\sqrt{p^{2}+m_{\mu}^{2}}. It follows that
There are several methods of solving this problem.
Method 1. We use the Doppler effect. In the system of the detector, the frequency of the photon is omega//h\omega / h. This frequency is related to the frequency omega^(')//h\omega^{\prime} / h of the photon in the system of the decaying particle, from which it is emitted, according to the formula for the Doppler shift as
where vv is the velocity of the decaying particle before it emits the photon. The frequency must be blueshifted, since the emitting particle moves toward the detector. Now, we have v=|p|//Ev=|\boldsymbol{p}| / E. Inserting this into the Doppler formula above gives, after some simplifications,
Method 2. Conservation of energy and momentum says that if the decaying particle emits a photon with 4-momentum k=(w,k)k=(w, \boldsymbol{k}) and a rest product (which can be several particles) of momentum p^(')=(E^('),p^('))p^{\prime}=\left(E^{\prime}, \boldsymbol{p}^{\prime}\right), then we have, in the rest frame of the decaying particle,
whence p^('2)=(p-k)^(2)=M^(2)-2p*kp^{\prime 2}=(p-k)^{2}=M^{2}-2 p \cdot k, where we have used k^(2)=0k^{2}=0. Inserting this into the first relation gives
{:(3.397)(M-omega^('))^(2)=M^(2)-2p*k+omega^('2):}\begin{equation*}
\left(M-\omega^{\prime}\right)^{2}=M^{2}-2 p \cdot k+\omega^{\prime 2} \tag{3.397}
\end{equation*}
If we solve for omega^(')\omega^{\prime}, we obtain omega^(')=p*k//M\omega^{\prime}=p \cdot k / M. By inserting p=(E,p)p=(E, \boldsymbol{p}) and k=(omega,k)k=(\omega, \boldsymbol{k}) with pp parallel to kk, we obtain the same answer as with Method 1 .
Method 3. The previous method suggests that one can solve the problem by studying the relativistic invariant p*kp \cdot k. In the rest frame of the decaying particle, its value is Momega^(')M \omega^{\prime}. In the frame of the detector, its value is
{:(3.398)p*k=E omega-p*k=omega(E-|p|):}\begin{equation*}
p \cdot k=E \omega-\boldsymbol{p} \cdot \boldsymbol{k}=\omega(E-|\boldsymbol{p}|) \tag{3.398}
\end{equation*}
Since the value of the invariant is independent of the frame, one finds Momega^(')=M \omega^{\prime}=omega(E-|p|)\omega(E-|\boldsymbol{p}|), which, after simplifications, leads to the same answer as with the previous two methods.
Method 4. Lastly, one can also simply make a Lorentz transformation of k=k=(omega^('),omega^('),0,0)\left(\omega^{\prime}, \omega^{\prime}, 0,0\right), which is 4 -wavevector of the photon in the rest frame of the decaying particle and where we have put the direction to the detector to coincide with the xx-axis to the detector system in which the particle moves with speed v=|p|//Ev=|\boldsymbol{p}| / E toward the detector. The detector then moves with speed -v-v relative to the particle. For the 0 -component, one then finds that
where gamma-=1//sqrt(1-v^(2))\gamma \equiv 1 / \sqrt{1-v^{2}}. Inserting the value of vv above gives, after simplifications, the same result as obtained by the other three methods.
1.84
In the rest frame of the positron, let the 4-momenta be p_(e)=(E_(e)//c,0,0,p)p_{e}=\left(E_{e} / c, 0,0, p\right) and p_(p)=(m_(e)c,0,0,0)p_{p}=\left(m_{e} c, 0,0,0\right) for the electron and the positron, respectively, where E_(e)E_{e} is the total energy of the electron, (0,0,p)(0,0, p) is the 3 -momentum of the electron, and m_(e)m_{e} is the mass of an electron or a positron. The 4-momenta of the photons are k_(1)=(omega_(1),omega_(1)sin phi,0,omega_(1)cos phi)k_{1}=\left(\omega_{1}, \omega_{1} \sin \phi, 0, \omega_{1} \cos \phi\right) and k_(2)=(omega_(2),-omega_(2)sin phi,0,omega_(2)cos phi)k_{2}=\left(\omega_{2},-\omega_{2} \sin \phi, 0, \omega_{2} \cos \phi\right), respectively. Conservation of 4-momentum gives
Hence, we find that omega_(1)=omega_(2)\omega_{1}=\omega_{2}.
a) From the relations above, we obtain the angle phi\phi as a function of the total energy of the electron E_(e)E_{e} as
b) In the nonrelativistic limit, i.e., E_(e)≃m_(e)c^(2)+p^(2)//(2m_(e))E_{e} \simeq m_{e} c^{2}+p^{2} /\left(2 m_{e}\right), with p≪m_(2)p \ll m_{2}, we find that
The energy for a photon is E=pc={p=(h)/( lambda)}=(hc)/(lambda)E=p c=\left\{p=\frac{h}{\lambda}\right\}=\frac{h c}{\lambda}. Using P_(1)=(h)/(lambda_(1))(1,1,0)P_{1}=\frac{h}{\lambda_{1}}(1,1,0), P_(2)=(h)/(lambda_(2))(1,-1,0)P_{2}=\frac{h}{\lambda_{2}}(1,-1,0), and P=(h)/( lambda)(1,cos theta,sin theta)P=\frac{h}{\lambda}(1, \cos \theta, \sin \theta), we obtain
First, define the Lorentz invariant total 4 -momentum squared of the pions as s=(p_(1)+p_(2))^(2)s=\left(p_{1}+p_{2}\right)^{2}. Using conservation of 4-momentum, i.e., P=p_(1)+p_(2)+kP=p_{1}+p_{2}+k, we can then calculate ss to be s=(P-k)^(2)=M(M-2omega)s=(P-k)^{2}=M(M-2 \omega). However, in the rest frame of the pions, we know that sqrts=2E\sqrt{s}=2 E, where E=sqrt(m^(2)+p^(2))E=\sqrt{m^{2}+p^{2}} and p^(2)=p_(1)^(2)=p_(2)^(2)p^{2}=p_{1}^{2}=p_{2}^{2}. Therefore, using the two expressions for ss, we obtain M^(2)-2M omega=4(m^(2)+p^(2))M^{2}-2 M \omega=4\left(m^{2}+p^{2}\right), which can be solved for pp to give
{:(3.411)p=(1)/(2)sqrt(M^(2)-2M omega-4m^(2)):}\begin{equation*}
p=\frac{1}{2} \sqrt{M^{2}-2 M \omega-4 m^{2}} \tag{3.411}
\end{equation*}
Finally, the speed of the pions relative to their center-of-mass frame is v=p//Ev=p / E, so inserting the expression for pp and E=(1)/(2)sqrts=(1)/(2)sqrt(M^(2)-2M omega)E=\frac{1}{2} \sqrt{s}=\frac{1}{2} \sqrt{M^{2}-2 M \omega}, we find that
{:(3.412)v=(p)/(E)=sqrt((M^(2)-2M omega-4m^(2))/(M^(2)-2M omega))=sqrt(1-(4m^(2))/(M(M-2omega))).:}\begin{equation*}
v=\frac{p}{E}=\sqrt{\frac{M^{2}-2 M \omega-4 m^{2}}{M^{2}-2 M \omega}}=\sqrt{1-\frac{4 m^{2}}{M(M-2 \omega)}} . \tag{3.412}
\end{equation*}
1.87
Before the decay, the Sigma^(0)\Sigma^{0} particle moves with 4-momentum ( E,pE, \boldsymbol{p} ) toward the detector. After the decay, the Lambda\Lambda particle moves toward the detector with 4-momentum ( E^('),p^(')E^{\prime}, \boldsymbol{p}^{\prime} ) and the photon with momentum ( omega^('),k^(')\omega^{\prime}, \boldsymbol{k}^{\prime} ).
a) The total energy of the Sigma^(0)\Sigma^{0} particle is given by
where E_(gamma)E_{\gamma} is the energy of the photon in the rest frame of the Sigma^(0)\Sigma^{0} particle. Solving for E_(gamma)E_{\gamma} results in
c) Since the Sigma^(0)\Sigma^{0} particle is moving straight toward the detector, the energy of the photon as registered by the detector will be given by the relativistic Doppler shift formula
1.88
a) In the center-of-mass system, we have by definition p_(e)+p_(p)=0\boldsymbol{p}_{e}+\boldsymbol{p}_{p}=\mathbf{0} and conservation of 3 -momentum then leads to p_(e)^(')+p_(p)^(')=0\boldsymbol{p}_{e}^{\prime}+\boldsymbol{p}_{p}^{\prime}=\mathbf{0}. Due to conservation of energy, we have for elastic scattering that |p_(e)|=|p_(p)|=|p_(e)^(')|=|p_(p)^(')|-=p\left|\boldsymbol{p}_{e}\right|=\left|\boldsymbol{p}_{p}\right|=\left|\boldsymbol{p}_{e}^{\prime}\right|=\left|\boldsymbol{p}_{p}^{\prime}\right| \equiv p. Therefore, we have E_(e)=E_(e)^(')E_{e}=E_{e}^{\prime}. Using these results, we find that t=(p_(e)-p_(e)^('))^(2)=t=\left(p_{e}-p_{e}^{\prime}\right)^{2}=(E_(e)-E_(e)^('))^(2)-(p_(e)-p_(e)^('))^(2)=-(p_(e)-p_(e)^('))^(2)\left(E_{e}-E_{e}^{\prime}\right)^{2}-\left(\boldsymbol{p}_{e}-\boldsymbol{p}_{e}^{\prime}\right)^{2}=-\left(\boldsymbol{p}_{e}-\boldsymbol{p}_{e}^{\prime}\right)^{2}, and thus, we obtain -t=(p_(e)-p_(e)^('))^(2)-t=\left(\boldsymbol{p}_{e}-\boldsymbol{p}_{e}^{\prime}\right)^{2}. Introducing the scattering angle theta\theta by p_(e)*p_(e)^(')=p^(2)cos theta\boldsymbol{p}_{e} \cdot \boldsymbol{p}_{e}^{\prime}=p^{2} \cos \theta, we find that
Thus, the result is -t=4p^(2)sin^(2)(theta//2)-t=4 p^{2} \sin ^{2}(\theta / 2).
b) Since conservation of 4-momentum holds, we also have p_(e)-p_(e)^(')=p_(p)^(')-p_(p)p_{e}-p_{e}^{\prime}=p_{p}^{\prime}-p_{p}, which means that t=(p_(p)^(')-p_(p))^(2)t=\left(p_{p}^{\prime}-p_{p}\right)^{2}. In the laboratory system, we have p_(p)=(m_(p),0)p_{p}=\left(m_{p}, \mathbf{0}\right) and p_(p)^(')=(m_(p)+T_(p)^('),p_(p))p_{p}^{\prime}=\left(m_{p}+T_{p}^{\prime}, \boldsymbol{p}_{p}\right). Therefore, we find that
Thus, the result is T_(p)^(')=-t//(2m_(p))T_{p}^{\prime}=-t /\left(2 m_{p}\right).
1.89
The energy of the pion before the collision is E=m_(pi)+TE=m_{\pi}+T. Now, s_(0)=(m_(pi)+m_(Delta))^(2)s_{0}=\left(m_{\pi}+m_{\Delta}\right)^{2} gives the minimal center-of-mass energy squared for production of the Delta\Delta. However, this is also given by s_(0)=((E,p)+(m_(p),0))^(2)=(E+m_(p))^(2)-p^(2)=(T+m_(pi)+:}s_{0}=\left((E, \boldsymbol{p})+\left(m_{p}, \mathbf{0}\right)\right)^{2}=\left(E+m_{p}\right)^{2}-\boldsymbol{p}^{2}=\left(T+m_{\pi}+\right.m_(p))^(2)-p^(2)\left.m_{p}\right)^{2}-\boldsymbol{p}^{2}. From E^(2)=(m+T)^(2)=m^(2)+p^(2)E^{2}=(m+T)^{2}=m^{2}+p^{2} follows that p^(2)=T^(2)+2mTp^{2}=T^{2}+2 m T. Thus, the kinetic energy TT of the pion required to create the Delta\Delta is given by
Consider the reaction pi^(+)+n longrightarrowK^(+)+Lambda\pi^{+}+n \longrightarrow K^{+}+\Lambda. Let theta=[pi^(+),K^(+)]=90^(@)\theta=\left[\pi^{+}, K^{+}\right]=90^{\circ}. Conservation of 4-momentum gives
The kinetic energy of the neutron is negligible (the kinetic energy of a thermal neutron is of the order 25 meV ), so both the neutron and the proton can be considered at rest before the reaction. By conservation of 4-momentum, we have the relation
If we let M=m_(p)+m_(n)M=m_{p}+m_{n}, then p_(p)+p_(n)=(M,0)p_{p}+p_{n}=(M, \mathbf{0}) and p_(gamma)=(E_(gamma),p_(gamma))p_{\gamma}=\left(E_{\gamma}, \boldsymbol{p}_{\gamma}\right) in the lab frame, while p_(d)=(M-B,0)p_{d}=(M-B, 0) in the rest frame of the deutron. It follows from the Lorentz invariance of the Minkowski product that
{:[M^(2)-2ME_(gamma)=M^(2)-2MB+B^(2)],[(3.444)quad=>quadE_(gamma)=B(1-(B)/(2M))=B(1-(B)/(2(m_(p)+m_(n))))]:}\begin{align*}
& M^{2}-2 M E_{\gamma}=M^{2}-2 M B+B^{2} \\
& \quad \Rightarrow \quad E_{\gamma}=B\left(1-\frac{B}{2 M}\right)=B\left(1-\frac{B}{2\left(m_{p}+m_{n}\right)}\right) \tag{3.444}
\end{align*}
The difference between the energies BB and E_(gamma)E_{\gamma} is due to the recoil energy of the deutron.
1.94
Let the 4 -momenta for the particles in the reaction be k,p_(H),p_(p)k, p_{\mathrm{H}}, p_{p}, and p_(e)p_{e}. Since 4 -momentum is conserved, we have k+p_(H)=p_(p)+p_(e)k+p_{\mathrm{H}}=p_{p}+p_{e}. From Lorentz invariance, we also have (k+p_(H))^(2)=(p_(p)+p_(e))^(2)\left(k+p_{\mathrm{H}}\right)^{2}=\left(p_{p}+p_{e}\right)^{2}. Now, since both sides of this relation are Lorentz invariants, we can calculate them in different inertial systems. In the system where H is at rest, we can take k=(omega,omega,0,0)k=(\omega, \omega, 0,0) and p_(H)=(m_(H),0,0,0)p_{\mathrm{H}}=\left(m_{\mathrm{H}}, 0,0,0\right). Thus, we obtain (k+p_(H))^(2)=2omegam_(H)+m_(H)^(2)\left(k+p_{\mathrm{H}}\right)^{2}=2 \omega m_{\mathrm{H}}+m_{\mathrm{H}}^{2}. For the final particles, we choose the center-ofmass frame, where the particles are at rest at threshold for the reaction to occur.
Thus, we find that (p_(p)+p_(e))^(2)=(m_(p)+m_(e))^(2)=(m_(H)+B)^(2)\left(p_{p}+p_{e}\right)^{2}=\left(m_{p}+m_{e}\right)^{2}=\left(m_{\mathrm{H}}+B\right)^{2}. Finally, combining the above expressions and solving for omega\omega, we obtain
In order for the collision of the two neutrinos to produce a Z^(0)Z^{0} boson, the center-ofmomentum energy must be equal to the Z^(0)Z^{0} mass, i.e.,
where m_(v)m_{v} is the neutrino mass, E_(k)E_{k} is the thermal energy of the CNB neutrino, EE is the energy of the UHE neutrino, and a\boldsymbol{a} and b\boldsymbol{b} are 3-vectors with modulus one (|a|=|b|=1)(|\boldsymbol{a}|=|\boldsymbol{b}|=1). Note that we have neglected the neutrino mass compared to the UHE neutrino energy. In general, we now need to take care if we obtain an expression, where terms including EE cancel (since the corrections of the small parameter m_(mu)//Em_{\mu} / E would then become the leading terms). However, this will not be the case in this problem.
Summing and then squaring the 4 -vectors results in
a) Conservation of 4 -momentum yields k+p=k^(')+p^(')k+p=k^{\prime}+p^{\prime}. We can remove the redundant information about p^(')p^{\prime} as follows
Using k^(2)=k^('2)=0k^{2}=k^{\prime 2}=0 and simplifying gives
{:(3.454)p*k-p*k^(')-k*k^(')=0:}\begin{equation*}
p \cdot k-p \cdot k^{\prime}-k \cdot k^{\prime}=0 \tag{3.454}
\end{equation*}
Inserting p=(m,0),k=(omega,k)p=(m, \mathbf{0}), k=(\omega, \boldsymbol{k}), and k^(')=(omega^('),k^('))k^{\prime}=\left(\omega^{\prime}, \boldsymbol{k}^{\prime}\right), this leads to
in the rest frame of the initial electron. We then use omega=2pi nu={nu=1//lambda}=2pi//lambda\omega=2 \pi \nu=\{\nu=1 / \lambda\}=2 \pi / \lambda and similar for omega^(')\omega^{\prime} to obtain the Compton formula, i.e.,
{:(3.457)s-=(p+k)^(2)=p^(2)+k^(2)+2p*k=m^(2)+2m omega:}\begin{equation*}
s \equiv(p+k)^{2}=p^{2}+k^{2}+2 p \cdot k=m^{2}+2 m \omega \tag{3.457}
\end{equation*}
On the other hand, due to conservation of 4-momentum, we also have, in the center-of-mass system (i.e., p^(')+k^(')=0\boldsymbol{p}^{\prime}+\boldsymbol{k}^{\prime}=\mathbf{0} ), the expression
where E^(')=sqrt(m^(2)+p^('2))E^{\prime}=\sqrt{m^{2}+\boldsymbol{p}^{\prime 2}} is the energy of the outgoing electron and p^('2)=k^('2)=omega^('2)\boldsymbol{p}^{\prime 2}=\boldsymbol{k}^{\prime 2}=\omega^{\prime 2}. Now, since ss is Lorentz invariant, equating the two expressions for ss and solving for omega^(')\omega^{\prime}, we obtain
Denoting the incoming 4-momenta of the electron and photon p_(e)p_{e} and p_(gamma)p_{\gamma}, respectively, and the corresponding outgoing quantities by k_(e)k_{e} and k_(gamma)k_{\gamma}, we have
in the laboratory frame, where we have oriented our coordinate system such that there is no momentum in the zz-direction either before or after the collision. Thus, we have also omitted the zz-components of the 4 -vectors. Conservation of 4-momentum yields
Furthermore, the total energy of the outgoing electron E_(e)E_{e} is given by E_(e)=m+E_{e}=m+omega-omega^(')\omega-\omega^{\prime} from the time component of the 4-momentum conservation (i.e., energy conservation). Since the kinetic energy is T_(e)=E_(e)-m=omega-omega^(')T_{e}=E_{e}-m=\omega-\omega^{\prime}, we insert our result for omega^(')\omega^{\prime} and simplify
where E_(0)E_{0} is the initial photon energy, vv is the initial speed of the electron, and EE is the sought energy of the photon after scattering. By conservation of 4-momentum, we find that
{:[m^(2)+2p*k-2(p+k)*k^(')=m^(2)quad=>quad p*k=(p+k)*k^(')],[(3.467)=>quad m gammaE_(0)(1-v)=m gamma E(1-v cos theta)+EE_(0)(1-cos theta)]:}\begin{align*}
& m^{2}+2 p \cdot k-2(p+k) \cdot k^{\prime}=m^{2} \quad \Rightarrow \quad p \cdot k=(p+k) \cdot k^{\prime} \\
& \Rightarrow \quad m \gamma E_{0}(1-v)=m \gamma E(1-v \cos \theta)+E E_{0}(1-\cos \theta) \tag{3.467}
\end{align*}
Figure 3.8 Setup of inverse Compton scattering gamma+e^(-)longrightarrow gamma+e^(-)\gamma+e^{-} \longrightarrow \gamma+e^{-}before and after scattering.
Consider the reaction mu^(+)longrightarrowe^(+)+v_(e)+ bar(v)_(mu)\mu^{+} \longrightarrow e^{+}+v_{e}+\bar{v}_{\mu}. Let p_(mu),p_(e),p_(v_(e))p_{\mu}, p_{e}, p_{v_{e}}, and p_( bar(v)_(mu))p_{\bar{v}_{\mu}} be the 4-momenta of the antimuon, positron, electron neutrino, and antimuon neutrino, respectively. Conservation of 4-momentum gives p_(mu)=p_(e)+p_(v_(e))+p_( bar(v)_(mu))p_{\mu}=p_{e}+p_{v_{e}}+p_{\bar{v}_{\mu}}, so that p_(mu)-p_(v_(e))=p_(e)+p_( bar(v)_(mu))-=Pp_{\mu}-p_{v_{e}}=p_{e}+p_{\bar{v}_{\mu}} \equiv P. Squaring both sides yields
All of the terms in this relation are Lorentz invariant and may be calculated in any inertial system. Calculating p_(e)*p_( bar(v)_(mu))p_{e} \cdot p_{\bar{v}_{\mu}} in the rest frame of the positron, it is easy to obtain p_(e)*p_( bar(v)_(mu)) >= m_(e)m_( bar(v)_(mu))p_{e} \cdot p_{\bar{v}_{\mu}} \geq m_{e} m_{\bar{v}_{\mu}}, which is negligible compared to m_(e)^(2)m_{e}^{2}. In the rest frame of the antimuon, we have p_(mu)=(m_(mu),0)p_{\mu}=\left(m_{\mu}, \mathbf{0}\right) and p_(v_(e))=(E,p)p_{v_{e}}=(E, \boldsymbol{p}), where |p|=sqrt(E^(2)-m_(v_(e))^(2))|\boldsymbol{p}|=\sqrt{E^{2}-m_{v_{e}}^{2}} and EE being the total energy of the electron neutrino. Inserting this into the equation above results in the inequality
{:(3.471)m_(mu)^(2)-2m_(mu)E >= m_(e)^(2)quad=>quad E <= (m_(mu)^(2)-m_(e)^(2))/(2m_(mu)):}\begin{equation*}
m_{\mu}^{2}-2 m_{\mu} E \geq m_{e}^{2} \quad \Rightarrow \quad E \leq \frac{m_{\mu}^{2}-m_{e}^{2}}{2 m_{\mu}} \tag{3.471}
\end{equation*}
Thus, the largest possible total energy of the electron neutrino in the rest frame of the antimuon is given by
where m_(mu)-=m_(mu^(-))=m_(mu^(+))m_{\mu} \equiv m_{\mu^{-}}=m_{\mu^{+}}and T_(mu)T_{\mu} is the kinetic energy of the mu^(+)\mu^{+}. In the rest frame of the mu^(-)\mu^{-}, the right-hand side becomes
Alternatively, one may realize that the maximal energy of the muons is given when the energy of the photon goes to zero. In that case, since the muons have identical masses, the total energy of the rho\rho-meson will be evenly divided to the total energy of the muons. The kinetic energy of one of the muons is then given by T_(mu)=E_(mu)-m_(mu)T_{\mu}=E_{\mu}-m_{\mu}, where E_(mu)=m_(rho)//2E_{\mu}=m_{\rho} / 2 is the total energy of one of the muons. Thus, the maximal kinetic energy that one of the muons can have in this decay in the rest frame of the rho\rho-meson is T_(mu)=m_(rho)//2-m_(mu)T_{\mu}=m_{\rho} / 2-m_{\mu}.
1.101
Giving all quantities in the rest frame of the ^(76)Ge{ }^{76} \mathrm{Ge}, we have that
where EE is the total energy of the two electrons and TT is their total kinetic energy, the quantity in which we are interested. Since the masses are invariant, we deduce that
is a function of E_(Se)E_{\mathrm{Se}} only. In order to maximize T,E_(Se)=sqrt(m_(Se)^(2)+p_(Se)^(2))T, E_{\mathrm{Se}}=\sqrt{m_{\mathrm{Se}}^{2}+p_{\mathrm{Se}}^{2}} must take its minimum allowed value, i.e., E_(Se)=m_(Se)E_{\mathrm{Se}}=m_{\mathrm{Se}}. It is necessary to check that this is kinematically allowed, which is the case since the momentum conservation is
solved by p_(1)=-p_(2)\boldsymbol{p}_{1}=-\boldsymbol{p}_{2}, which leaves the energy of the electrons as a free parameter, which may be adjusted to solve the energy conservation. It follows that
The minimal energy of the electrons is instead obtained when the ^(76)Se{ }^{76} \mathrm{Se} obtains its maximal energy. Conservation of 4-momentum gives
which is typically a very small number since the the difference between the masses of the nuclei is relatively small. Thus, the electron spectrum for the neutrinoless double beta decay is very peaked. This is in sharp contrast to the case of the more common double beta decay ( X longrightarrow Y+2e^(-)+2 bar(v)_(e)X \longrightarrow Y+2 e^{-}+2 \bar{v}_{e} ), where the electron spectrum is continuous and broad due to the possibility of the neutrinos taking some of the energy.
1.102
According to the conservation of 4-momentum, the 4-momentum of the new particle phi\phi must be given by
In the laboratory (lab) system: p_("lab ")=(E+m_(p),p,0,0)p_{\text {lab }}=\left(E+m_{p}, p, 0,0\right),
In the center-of-mass (CM) system: p_(CM)=(E_(**),0,0,0)p_{\mathrm{CM}}=\left(E_{*}, 0,0,0\right),
where EE and pp are the energy and the momentum of the incoming proton, respectively, E_(**)E_{*} is the energy in the CM system, and m_(p)m_{p} is the rest energy (mass) for a proton (or antiproton).
The 4-momentum squared is an invariant, and thus, the same in the two systems. Therefore, we have
The reaction is therefore not possible at this kinetic energy.
b) Let the 4-momentum of the eta\eta be p_(eta)=(m_(eta)+T,p_(eta))p_{\eta}=\left(m_{\eta}+T, \boldsymbol{p}_{\eta}\right), where TT is the kinetic energy and p_(eta)\boldsymbol{p}_{\eta} is the 3-momentum in the rest frame of the nucleons. Squaring the relation from conservation of 4 -momentum now results in
{:(3.502)9m^(2)+4mT_(p)=[(3m,0)+p_(eta)]^(2)=9m^(2)+6m(m_(eta)+T)+m_(eta)^(2):}\begin{equation*}
9 m^{2}+4 m T_{p}=\left[(3 m, 0)+p_{\eta}\right]^{2}=9 m^{2}+6 m\left(m_{\eta}+T\right)+m_{\eta}^{2} \tag{3.502}
\end{equation*}
where the term in brackets only appears in the 1pi1 \pi reaction. Squaring this relation gives us p_(v)^(2)+2p_(v)*p_(X)+p_(X)^(2)=2p_(v)*p_(X)+m_(X)^(2)=(p_(mu)+p_(Y)[+p_(pi)])^(2) >= (m_(mu)+m_(Y)[+m_(pi)])^(2)p_{v}^{2}+2 p_{v} \cdot p_{X}+p_{X}^{2}=2 p_{v} \cdot p_{X}+m_{X}^{2}=\left(p_{\mu}+p_{Y}\left[+p_{\pi}\right]\right)^{2} \geq\left(m_{\mu}+m_{Y}\left[+m_{\pi}\right]\right)^{2}.
In the rest frame of XX, the product p_(v)*p_(X)p_{v} \cdot p_{X} evaluates to
We denote the incoming momenta p_(1)p_{1} and p_(2)p_{2}, respectively, while we call the outgoing momenta k_(i),i=1,2,3,4k_{i}, i=1,2,3,4. Since all of the particles have the same mass, we have the relation p_(i)^(2)=k_(i)^(2)=m^(2)p_{i}^{2}=k_{i}^{2}=m^{2}. By conservation of 4-momentum, we have
Squaring this relation and using the inequality A*B >= sqrt(A^(2)B^(2))A \cdot B \geq \sqrt{A^{2} B^{2}} for any timelike vectors AA and BB, we find that
where EE is the total energy of the other electron. It follows that
{:(3.514)mE=7m^(2)=>E=7m:}\begin{equation*}
m E=7 m^{2} \Rightarrow E=7 m \tag{3.514}
\end{equation*}
On the other hand, in the center-of-momentum frame, the total energy squared is the square of the total 4-momentum. Thus, the total center-of-mass (CM) energy is given by
{:(3.515)E_(CM)=sqrt((4m)^(2))=4m:}\begin{equation*}
E_{\mathrm{CM}}=\sqrt{(4 m)^{2}}=4 m \tag{3.515}
\end{equation*}
at threshold. The ratio E//E_(CM)E / E_{\mathrm{CM}} is therefore 7//47 / 4. Note that the total energy in the frame where one of the electrons is at rest is E+m=8mE+m=8 m.
where p_(i)p_{i} are the 4-momenta of the incoming protons, p_(i)^(')p_{i}^{\prime} are the 4 -momenta of the outgoing protons, and p_(K)^(')p_{K}^{\prime} is the 4 -momentum of the kaon. By squaring this relation, we obtain
{:(3.527)T >= (1+(m_(chi))/(m_(p)))delta+(delta^(2))/(2m_(p)):}\begin{equation*}
T \geq\left(1+\frac{m_{\chi}}{m_{p}}\right) \delta+\frac{\delta^{2}}{2 m_{p}} \tag{3.527}
\end{equation*}
In the limit delta≪m_(chi)\delta \ll m_{\chi}, we can neglect the delta^(2)\delta^{2} term and obtain
{:(3.528)T >= (1+(m_(chi))/(m_(p)))delta:}\begin{equation*}
T \geq\left(1+\frac{m_{\chi}}{m_{p}}\right) \delta \tag{3.528}
\end{equation*}
As long as delta//T≪m_(chi)//m_(p)\delta / T \ll m_{\chi} / m_{p}, the threshold energy TT will be in the classical limit and given by T=m_(chi)v^(2)//2T=m_{\chi} v^{2} / 2. This leads to
where mu-=m_(p)m_(chi)//(m_(p)+m_(chi))\mu \equiv m_{p} m_{\chi} /\left(m_{p}+m_{\chi}\right) is the reduced mass of the proton- chi\chi system.
1.110
Since, for a particle of mass m,p^(mu)=(E,p)^(mu)=mV^(mu)=m gamma(1,v)^(mu)m, p^{\mu}=(E, \boldsymbol{p})^{\mu}=m V^{\mu}=m \gamma(1, \boldsymbol{v})^{\mu}, it follows that
where E_("in ")E_{\text {in }} is the initial total energy, and E_("acc ")E_{\text {acc }} is the energy added by acceleration. Since the particle was initially at rest, we have E_("in ")=mc^(2)E_{\text {in }}=m c^{2}, where mm is the mass of the particle. From the relation
where TT is the kinetic energy, it immediately follows that T=E_("acc. ")T=E_{\text {acc. }}. The energy added by the accelerator is given by
{:(3.536)E_(acc)=Q Delta U:}\begin{equation*}
E_{\mathrm{acc}}=Q \Delta U \tag{3.536}
\end{equation*}
where Q=eQ=e is the charge of the particle and Delta U\Delta U is the difference in the electric potential at the start and end of the acceleration. Since the electric field is known, we can easily compute Delta U\Delta U as
{:(3.544)t=(1)/(ceE^('))int_(mc^(2))^(mc^(2)+E_(acc))(EdE)/(sqrt(E^(2)-m^(2)c^(4))):}\begin{equation*}
t=\frac{1}{c e E^{\prime}} \int_{m c^{2}}^{m c^{2}+E_{\mathrm{acc}}} \frac{E d E}{\sqrt{E^{2}-m^{2} c^{4}}} \tag{3.544}
\end{equation*}
when inserted into the integral. A new change of variables to alpha=E^(2)\alpha=E^{2} results in the formula
{:[t=(1)/(ceE^('))int_(m^(2)c^(4))^((mc^(2)+E_(acc))^(2))(d alpha)/(2sqrt(alpha-m^(2)c^(4)))=(1)/(ceE^('))[sqrt(alpha-m^(2)c^(4))]_(m^(2)c^(4))^((mc^(2)+E_(acc))^(2))],[(3.545){:=(1)/(ceE^('))sqrt(E_(acc)(2mc^(2)+E_(acc)))=(1)/(ceE^(')),sqrt(eE^(')L(2mc^(2)+eE^(')L:}))]:}\begin{align*}
t & =\frac{1}{c e E^{\prime}} \int_{m^{2} c^{4}}^{\left(m c^{2}+E_{\mathrm{acc}}\right)^{2}} \frac{d \alpha}{2 \sqrt{\alpha-m^{2} c^{4}}}=\frac{1}{c e E^{\prime}}\left[\sqrt{\alpha-m^{2} c^{4}}\right]_{m^{2} c^{4}}^{\left(m c^{2}+E_{\mathrm{acc}}\right)^{2}} \\
& \left.=\frac{1}{c e E^{\prime}} \sqrt{E_{\mathrm{acc}}\left(2 m c^{2}+E_{\mathrm{acc}}\right)}=\frac{1}{c e E^{\prime}}, \sqrt{e E^{\prime} L\left(2 m c^{2}+e E^{\prime} L\right.}\right) \tag{3.545}
\end{align*}
1.112
Let A^(mu)=A^(mu)(x)A^{\mu}=A^{\mu}(x) be the solution to ◻A^(mu)(x)=0\square A^{\mu}(x)=0. Calculate ◻^(')A^('mu)(x^('))\square^{\prime} A^{\prime \mu}\left(x^{\prime}\right), where x^(')=Lambda xx^{\prime}=\Lambda x and A^('mu)(x^('))=Lambda^(mu)_(v)A^(v)(x)A^{\prime \mu}\left(x^{\prime}\right)=\Lambda^{\mu}{ }_{v} A^{v}(x). An explicit calculation yields
Inserting the definition A_(mu)^(')=A_(mu)+del_(mu)psiA_{\mu}^{\prime}=A_{\mu}+\partial_{\mu} \psi into F_(mu nu)^(')=del_(mu)A_(nu)^(')-del_(nu)A_(mu)^(')F_{\mu \nu}^{\prime}=\partial_{\mu} A_{\nu}^{\prime}-\partial_{\nu} A_{\mu}^{\prime}, we obtain
such that x^(')=Lambda xx^{\prime}=\Lambda x.
a) The observer in K^(')K^{\prime} must measure the stick simultaneously in both endpoints. This means at time x^('0)=0x^{\prime 0}=0 in his/her coordinate system. Without loss of generality, we can put one of the endpoints of the stick at the origin in K^(')K^{\prime}. Thus, we have (0,0,0)(0,0,0) in K^(')K^{\prime} at time x^('0)=0x^{\prime 0}=0 and (ℓ,0,0)(\ell, 0,0) in K^(')K^{\prime} at time x^('0)=0x^{\prime 0}=0. Therefore, it holds that Deltax^('0)=0=Deltax^(0)cosh theta-ℓsinh theta\Delta x^{\prime 0}=0=\Delta x^{0} \cosh \theta-\ell \sinh \theta, which leads to
The relation cosh theta=1//sqrt(1-v^(2)//c^(2))\cosh \theta=1 / \sqrt{1-v^{2} / c^{2}} implies that ℓ^(')=ℓsqrt(1-v^(2)//c^(2))\ell^{\prime}=\ell \sqrt{1-v^{2} / c^{2}}. Thus, the result is
b) The electric and magnetic fields E=(0,0,E)\mathbf{E}=(0,0, E) and B=(0,0,0)\mathbf{B}=(0,0,0) in KK implies that the electromagnetic field strength tensor in KK is
Therefore, E^('1)=E^('2)=0,E^('3)=E cosh theta,B^('1)=B^('3)=0E^{\prime 1}=E^{\prime 2}=0, E^{\prime 3}=E \cosh \theta, B^{\prime 1}=B^{\prime 3}=0, and B^('2)=(E)/(c)sinh thetaB^{\prime 2}=\frac{E}{c} \sinh \theta. Using the relations cosh theta=gamma(v)\cosh \theta=\gamma(v) and sinh theta=(v)/(c)gamma(v)\sinh \theta=\frac{v}{c} \gamma(v), where gamma(v)=(1)/(sqrt(1-((v)/(c))^(2)))\gamma(v)=\frac{1}{\sqrt{1-\left(\frac{v}{c}\right)^{2}}}, we obtain the electric and magnetic fields in K^(')K^{\prime} as
Consider the electric and magnetic fields in KK, i.e., E=(E^(1),E^(2),E^(3))\mathbf{E}=\left(E^{1}, E^{2}, E^{3}\right) and B=0\mathbf{B}=\mathbf{0}. Thus, the corresponding electromagnetic field strength tensor in K^(')K^{\prime} is given by F^(')=Lambda FLambda^(T)F^{\prime}=\Lambda F \Lambda^{T}, where F=(F^(mu nu))=([0,-E^(1),-E^(2),-E^(3)],[E^(1),0,0,0],[E^(2),0,0,0],[E^(3),0,0,0]),quad Lambda=(Lambda^(mu)_(nu))≃([1,-beta,0,0],[-beta,1,0,0],[0,0,1,0],[0,0,0,1])F=\left(F^{\mu \nu}\right)=\left(\begin{array}{cccc}0 & -E^{1} & -E^{2} & -E^{3} \\ E^{1} & 0 & 0 & 0 \\ E^{2} & 0 & 0 & 0 \\ E^{3} & 0 & 0 & 0\end{array}\right), \quad \Lambda=\left(\Lambda^{\mu}{ }_{\nu}\right) \simeq\left(\begin{array}{cccc}1 & -\beta & 0 & 0 \\ -\beta & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1\end{array}\right),
are the electromagnetic field strength tensor in KK and the Lorentz transformation in the x^(1)x^{1}-direction with velocity vv, respectively. Therefore, we find that
In addition, the electric field in K^(')K^{\prime} is E^(')=(E^('1),E^('2),E^('3))=(E^(1),E^(2),E^(3))=E\mathbf{E}^{\prime}=\left(E^{\prime 1}, E^{\prime 2}, E^{\prime 3}\right)=\left(E^{1}, E^{2}, E^{3}\right)=\mathbf{E}. Clearly, B^(')\mathbf{B}^{\prime} is perpendicular to both the x^(1)x^{1}-axis (i.e., the direction of the velocity v=-ve_(1)\mathbf{v}=-v \mathbf{e}_{1} of KK in K^(')K^{\prime} ) and E^(')\mathbf{E}^{\prime}, since it holds that
It holds that B^(')*Eprop(vxxE)*E=0\mathbf{B}^{\prime} \cdot \mathbf{E} \propto(\mathbf{v} \times \mathbf{E}) \cdot \mathbf{E}=0 and B^(')*vprop(vxxE)*v=0\mathbf{B}^{\prime} \cdot \mathbf{v} \propto(\mathbf{v} \times \mathbf{E}) \cdot \mathbf{v}=0.
1.116
Consider the electric and magnetic fields in a coordinate system of KK, i.e., E=\mathbf{E}=(0,1,0)(0,1,0) and B=0\mathbf{B}=\mathbf{0}, which means that the electromagnetic field components are
Inserting the electromagnetic field components in the coordinate system of KK into the Lorentz transformation corresponding to a velocity boost v_(x)v_{x} in the x^(1)x^{1}-direction, namely
where gamma(v)-=(1)/(sqrt(1-v^(2)//c^(2)))\gamma(v) \equiv \frac{1}{\sqrt{1-v^{2} / c^{2}}}. Similarly, with a velocity boost v_(y)v_{y} in the x^('2)x^{\prime 2}-direction, the electromagnetic field components in the coordinate system of K^('')K^{\prime \prime} are given by
where quad cosh theta=gamma(v)-=1//sqrt(1-v^(2)//c^(2)),quad sinh theta=v gamma(v)//c,quad\quad \cosh \theta=\gamma(v) \equiv 1 / \sqrt{1-v^{2} / c^{2}}, \quad \sinh \theta=v \gamma(v) / c, \quad and quad r=\quad r=sqrt((x^(1))^(2)+(x^(2))^(2)+(x^(3))^(2))\sqrt{\left(x^{1}\right)^{2}+\left(x^{2}\right)^{2}+\left(x^{3}\right)^{2}}.
Note that in order to compute the electromagnetic field strengths at the point x^(')x^{\prime}, we have to write the xx-coordinates on the right-hand side of the equations in terms of the new x^(')x^{\prime}-coordinates. After doing so, we obtain
for the electromagnetic field of a point charge moving along the negative x^('1)x^{\prime 1}-axis with velocity vv, i.e., we have an electric current in the negative x^('1)x^{\prime 1}-direction.
1.118
Maxwell's equations describe how sources (charges and currents) give rise to electric and magnetic fields, whereas the Lorentz force law describes how the field strengths determine the trajectory of a moving test particle with rest mass mm and electric charge qq. Let us parametrize the trajectory of the particle as x=x(s)x=x(s), where ss is the proper time parameter. The Lorentz force law is then given by
{:(3.592)mc^(2)x^(¨)^(mu)(s)=qx^(˙)_(v)(s)F^(mu v)(x(s)):}\begin{equation*}
m c^{2} \ddot{x}^{\mu}(s)=q \dot{x}_{v}(s) F^{\mu v}(x(s)) \tag{3.592}
\end{equation*}
which is covariant under Lorentz transformations, i.e., both sides of the equation transform as 4 -vectors, and where F^(mu nu)F^{\mu \nu} is the electromagnetic field strength tensor. In order to understand the physical meaning of the Lorentz force law, we will first replace proper time derivatives by ordinary time derivatives, using x^(0)=ctx^{0}=c t. For the time derivatives of the spatial components of x=x(s)x=x(s), we have
Now, what is mux^(˙)^(0)m u \dot{x}^{0} ? The relativistic 4-momentum is defined as p^(mu)=mcx^(˙)^(mu)p^{\mu}=m c \dot{x}^{\mu}, so we have p=mcx^(˙)=mux^(˙)^(0)\boldsymbol{p}=m c \dot{\boldsymbol{x}}=m \boldsymbol{u} \dot{x}^{0} ( note that {:p!=mu)\left.\boldsymbol{p} \neq m \boldsymbol{u}\right), using x^(˙)=(1)/(c)ux^(˙)^(0)\dot{\boldsymbol{x}}=\frac{1}{c} \boldsymbol{u} \dot{x}^{0}. Thus, the Lorentz force law for the spatial components is given by
{:(3.596)(dp)/(dt)=q(E+u xx B)",":}\begin{equation*}
\frac{d p}{d t}=q(\boldsymbol{E}+\boldsymbol{u} \times \boldsymbol{B}), \tag{3.596}
\end{equation*}
where pp is the relativistic 3-momentum and E\boldsymbol{E} and B\boldsymbol{B} are the electric and magnetic fields, respectively. Furthermore, what is x^(˙)^(0)\dot{x}^{0} ? The proper time parameter ss is defined such that x^(˙)^(2)=(x^(˙)^(0))^(2)-x^(˙)^(2)=1\dot{x}^{2}=\left(\dot{x}^{0}\right)^{2}-\dot{\boldsymbol{x}}^{2}=1. Combining this with cx^(˙)=ux^(˙)^(0)c \dot{x}=u \dot{x}^{0}, we deduce
Assuming E\boldsymbol{E} is a constant electric field and B=0\boldsymbol{B}=\mathbf{0} as well as multiplying both sides with drd \boldsymbol{r}, we find that
{:(3.599)(dp)/(dt)*dr=qE*dr=qE*udt:}\begin{equation*}
\frac{d p}{d t} \cdot d \boldsymbol{r}=q \boldsymbol{E} \cdot d \boldsymbol{r}=q \boldsymbol{E} \cdot \boldsymbol{u} d t \tag{3.599}
\end{equation*}
Inserting x^(˙)^(0)=1//sqrt(1-u^(2)//c^(2))\dot{x}^{0}=1 / \sqrt{1-u^{2} / c^{2}} and using straightforward differential calculus, gives for the left-hand side
Integrating both sides from the origin, where u=0u=0, to the displacement rr, where the velocity momentarily is uu, we find that
{:[int_(u(0)=0)^(u(r))(dp)/(dt)*dr=int_(r=0)^(r)qE*dr=>(mc^(2))/(sqrt(1-u^(2)//c^(2)))-mc^(2)=qEr],[(3.602)=>(1)/(sqrt(1-u^(2)//c^(2)))=1+(qEr)/(mc^(2))]:}\begin{align*}
\int_{u(0)=0}^{u(r)} \frac{d p}{d t} \cdot d \boldsymbol{r}=\int_{r=0}^{r} q \boldsymbol{E} \cdot d \boldsymbol{r} & \Rightarrow \frac{m c^{2}}{\sqrt{1-u^{2} / c^{2}}}-m c^{2}=q E r \\
& \Rightarrow \frac{1}{\sqrt{1-u^{2} / c^{2}}}=1+\frac{q E r}{m c^{2}} \tag{3.602}
\end{align*}
since the electric field E\boldsymbol{E} is constant along the trajectory. Finally, introducing x-=x \equiv1+(qEr)/(mc^(2))1+\frac{q E r}{m c^{2}} and solving for uu, we obtain
{:(3.603)u(r)=csqrt(1-x^(-2))=csqrt(1-(1+(qEr)/(mc^(2)))^(-2)):}\begin{equation*}
u(r)=c \sqrt{1-x^{-2}}=c \sqrt{1-\left(1+\frac{q E r}{m c^{2}}\right)^{-2}} \tag{3.603}
\end{equation*}
which is the velocity u(r)u(r) of the particle as a function of the displacement rr from the origin along the direction of motion.
An alternative solution is to note that the electric field in the momentary rest frame K^(@)\stackrel{\circ}{K} of the electron is, according to the transformation equations for the field tensor, equal to the electric field in the laboratory system. Using units with c=1c=1, the acceleration of the electron relative to K^(@)\stackrel{\circ}{K} is therefore a^(@)=eE//m_(0)\stackrel{\circ}{a}=e E / m_{0}, where ee is the electron charge and m_(0)m_{0} is the electron rest mass. We obtain the answer by changing g rarr eE//m_(0)g \rightarrow e E / m_{0} in the formula [see Problem 1.41, Eq. (3.166)]
{:(3.605)x=(m_(0))/(eE)[sqrt(1+((eEt)/(m_(0)))^(2))-1]:}\begin{equation*}
x=\frac{m_{0}}{e E}\left[\sqrt{1+\left(\frac{e E t}{m_{0}}\right)^{2}}-1\right] \tag{3.605}
\end{equation*}
We now want to calculate the velocity as a function of the displacement, we multiply the Lorentz force f=e(E+u xx B)\boldsymbol{f}=e(\boldsymbol{E}+\boldsymbol{u} \times \boldsymbol{B}) by u\boldsymbol{u} and use that the change in kinetic energy is dT=f*drd T=\boldsymbol{f} \cdot d \boldsymbol{r}. We then introduce the electrostatic potential Phi\Phi(E=-grad Phi)(\boldsymbol{E}=-\nabla \Phi) :
{:(3.607)T=m_(0)(gamma(u)-1)=-e Delta Phi=eEr:}\begin{equation*}
T=m_{0}(\gamma(u)-1)=-e \Delta \Phi=e E r \tag{3.607}
\end{equation*}
Figure 3.9 A straight uncharged conductor with current II as viewed in the inertial systems KK and K^(')K^{\prime}. Note that K^(')K^{\prime} is moving with velocity vv relative to KK along the x^(1)x^{1}-axis.
i.e.,
{:(3.608)u(r)=sqrt(1-(1+(eEr)/(m_(0)))^(-2)):}\begin{equation*}
u(r)=\sqrt{1-\left(1+\frac{e E r}{m_{0}}\right)^{-2}} \tag{3.608}
\end{equation*}
1.119
Introduce KK and K^(')K^{\prime} according to Figure 3.9.
a) The components of the electromagnetic field in KK at the point P=(0,0,r)P=(0,0, r) are given by
b) The current-density 4 -vector in the conductor has the components J^(mu)=J^{\mu}=(0,cmu_(0)I//A,0,0)\left(0, c \mu_{0} I / A, 0,0\right) in KK and J^('mu)=(rho^(')//epsilon_(0),cmu_(0)j^('1),cmu_(0)j^('2),cmu_(0)j^('3))J^{\prime \mu}=\left(\rho^{\prime} / \epsilon_{0}, c \mu_{0} j^{\prime 1}, c \mu_{0} j^{\prime 2}, c \mu_{0} j^{\prime 3}\right) in K^(')K^{\prime}, where J^('mu)=Lambda_(nu)^(mu)J^(v),LambdaJ^{\prime \mu}=\Lambda_{\nu}^{\mu} J^{v}, \Lambda being the Lorentz transformation from KK to K^(')K^{\prime} and AA the crosssectional area of the conductor. We then obtain
{:(3.612)rho^(')=-(v gamma(v))/(c^(2))(I)/(A)quad" and "quadj^('1)=gamma(v)(I)/(A):}\begin{equation*}
\rho^{\prime}=-\frac{v \gamma(v)}{c^{2}} \frac{I}{A} \quad \text { and } \quad j^{\prime 1}=\gamma(v) \frac{I}{A} \tag{3.612}
\end{equation*}
Since the cross-sectional area A^(')A^{\prime} relative to K^(')K^{\prime} is AA, the current relative to K^(')K^{\prime} becomes
{:(3.613)I^(')=A^(')j^('1)=gamma(v)I:}\begin{equation*}
I^{\prime}=A^{\prime} j^{\prime 1}=\gamma(v) I \tag{3.613}
\end{equation*}
Now, rho^(')\rho^{\prime} and I^(')I^{\prime} generate the components of the electromagnetic field in K^(')K^{\prime} as
Inserting the definition of the electromagnetic field strength tensor, i.e., F^(mu nu)=F^{\mu \nu}=del^(mu)A^(nu)-del^(nu)A^(mu)\partial^{\mu} A^{\nu}-\partial^{\nu} A^{\mu}, into Maxwell's equations del_(mu)F^(mu nu)=J^(nu)\partial_{\mu} F^{\mu \nu}=J^{\nu}, one obtains
This is the simple form of Maxwell's equations and is a wave equation for AA with source term JJ. Assuming that J=0J=0, which implies that J^(v)=0J^{v}=0, one finds ◻A^(v)=0\square A^{v}=0. A useful formula is given by
{:(3.617)del^(mu)A^(nu)=del^(mu)(epsi^(nu)e^(ik*x))=epsi^(nu)del^(mu)e^(ik*x)=epsi^(nu)e^(ik*x)ik^(mu)=ik^(mu)epsi^(v)e^(ik*x)=ik^(mu)A^(nu):}\begin{equation*}
\partial^{\mu} A^{\nu}=\partial^{\mu}\left(\varepsilon^{\nu} e^{i k \cdot x}\right)=\varepsilon^{\nu} \partial^{\mu} e^{i k \cdot x}=\varepsilon^{\nu} e^{i k \cdot x} i k^{\mu}=i k^{\mu} \varepsilon^{v} e^{i k \cdot x}=i k^{\mu} A^{\nu} \tag{3.617}
\end{equation*}
Using Eq. (3.617), one obtains the electric and magnetic field components as
Multiplying Eqs. (3.618) and (3.619) with k^(i)k^{i}, one finds that
{:[E*k=E^(i)k^(i)=i(k^(i)epsi^(0)-k^(0)epsi^(i))e^(ik*x)k^(i)=i(k^(i)k^(i)epsi^(0)-k^(0)k^(i)epsi^(i))e^(ik*x)],[(3.620)=i(k^(2)epsi^(0)-k^(0)k*epsi)e^(ik*x)],[B*k=B^(i)k^(i)=epsilon^(ijk)ik^(j)A^(k)k^(i)=iepsilon^(ijk)k^(i)k^(j)A^(k)],[(3.621)={epsilon^(ijk)" is antisymmetric and "k^(i)k^(j)" is symmetric "}=0]:}\begin{align*}
\boldsymbol{E} \cdot \boldsymbol{k} & =E^{i} k^{i}=i\left(k^{i} \varepsilon^{0}-k^{0} \varepsilon^{i}\right) e^{i k \cdot x} k^{i}=i\left(k^{i} k^{i} \varepsilon^{0}-k^{0} k^{i} \varepsilon^{i}\right) e^{i k \cdot x} \\
& =i\left(\boldsymbol{k}^{2} \varepsilon^{0}-k^{0} k \cdot \varepsilon\right) e^{i k \cdot x} \tag{3.620}\\
\boldsymbol{B} \cdot \boldsymbol{k} & =B^{i} k^{i}=\epsilon^{i j k} i k^{j} A^{k} k^{i}=i \epsilon^{i j k} k^{i} k^{j} A^{k} \\
& =\left\{\epsilon^{i j k} \text { is antisymmetric and } k^{i} k^{j} \text { is symmetric }\right\}=0 \tag{3.621}
\end{align*}
Multiplying Eq. (3.617) with eta_(mu nu)\eta_{\mu \nu}, one has
{:(3.622)del_(mu)A^(mu)=eta_(mu v)del^(mu)A^(v)=eta_(mu v)ik^(mu)A^(v)=ik_(mu)A^(mu)=ik_(mu)epsi^(mu)e^(ik*x)=i(k_(0)epsi^(0)-k*epsi)e^(ik*x):}\begin{equation*}
\partial_{\mu} A^{\mu}=\eta_{\mu v} \partial^{\mu} A^{v}=\eta_{\mu v} i k^{\mu} A^{v}=i k_{\mu} A^{\mu}=i k_{\mu} \varepsilon^{\mu} e^{i k \cdot x}=i\left(k_{0} \varepsilon^{0}-k \cdot \varepsilon\right) e^{i k \cdot x} \tag{3.622}
\end{equation*}
but del_(mu)A^(mu)=0\partial_{\mu} A^{\mu}=0 (Lorenz gauge), so i(k_(0)epsi^(0)-k*epsi)e^(ik*x)=0i\left(k_{0} \varepsilon^{0}-k \cdot \varepsilon\right) e^{i k \cdot x}=0, i.e.,
Inserting the expression for the free electromagnetic plane wave A^(mu)(x)=epsi^(mu)e^(ik*x)A^{\mu}(x)=\varepsilon^{\mu} e^{i k \cdot x} into the definition of the electromagnetic field strength tensor F^(mu nu)=del^(mu)A^(nu)-del^(nu)A^(mu)F^{\mu \nu}=\partial^{\mu} A^{\nu}-\partial^{\nu} A^{\mu}, one obtains
{:(3.628)F^(mu v)=ik^(mu)epsi^(v)e^(ik*x)-ik^(v)epsi^(mu)e^(ik*x)=i(k^(mu)A^(v)-k^(nu)A^(mu)):}\begin{equation*}
F^{\mu v}=i k^{\mu} \varepsilon^{v} e^{i k \cdot x}-i k^{v} \varepsilon^{\mu} e^{i k \cdot x}=i\left(k^{\mu} A^{v}-k^{\nu} A^{\mu}\right) \tag{3.628}
\end{equation*}
Now, Maxwell's equations del_(mu)F^(mu nu)=J^(nu)\partial_{\mu} F^{\mu \nu}=J^{\nu} expressed in the 4-vector potential AA when the 4 -current J=0J=0, i.e., the free case, are ◻A^(nu)-del^(v)del*A=0\square A^{\nu}-\partial^{v} \partial \cdot A=0 and in kk-space
{:(3.630)-k^(2)A^(v)+k^(v)k*A=0:}\begin{equation*}
-k^{2} A^{v}+k^{v} k \cdot A=0 \tag{3.630}
\end{equation*}
Therefore, multiplying the above equation with A_(nu)A_{\nu}, one has
i.e., the invariant F_(mu nu)F^(mu nu)F_{\mu \nu} F^{\mu \nu} is zero. Finally, by writing out the invariant epsilon_(mu nu omega lambda)F^(mu nu)F^(omega lambda)\epsilon_{\mu \nu \omega \lambda} F^{\mu \nu} F^{\omega \lambda} explicitly, one finds that
{:[epsilon_(mu nu omega lambda)F^(mu nu)F^(omega lambda)=-epsilon_(mu nu omega lambda)(k^(mu)epsilon^(nu)-k^(nu)epsilon^(mu))(k^(omega)epsilon^(lambda)-k^(lambda)epsilon^(omega))e^(2ik*x)],[=-epsilon_(mu nu omega lambda)(k^(mu)k^(omega)epsilon^(nu)epsilon^(lambda)-k^(mu)k^(lambda)epsilon^(nu)epsilon^(omega)-k^(nu)k^(omega)epsilon^(mu)epsilon^(lambda)+k^(nu)k^(lambda)epsilon^(mu)epsilon^(omega))e^(2ik*x)],[(3.633)=0]:}\begin{align*}
\epsilon_{\mu \nu \omega \lambda} F^{\mu \nu} F^{\omega \lambda} & =-\epsilon_{\mu \nu \omega \lambda}\left(k^{\mu} \epsilon^{\nu}-k^{\nu} \epsilon^{\mu}\right)\left(k^{\omega} \epsilon^{\lambda}-k^{\lambda} \epsilon^{\omega}\right) e^{2 i k \cdot x} \\
& =-\epsilon_{\mu \nu \omega \lambda}\left(k^{\mu} k^{\omega} \epsilon^{\nu} \epsilon^{\lambda}-k^{\mu} k^{\lambda} \epsilon^{\nu} \epsilon^{\omega}-k^{\nu} k^{\omega} \epsilon^{\mu} \epsilon^{\lambda}+k^{\nu} k^{\lambda} \epsilon^{\mu} \epsilon^{\omega}\right) e^{2 i k \cdot x} \\
& =0 \tag{3.633}
\end{align*}
since epsilon_(mu nu omega lambda)\epsilon_{\mu \nu \omega \lambda} is totally antisymmetric and each term inside the parenthesis has two pairs of symmetric indices, i.e., the invariant epsilon_(mu nu omega lambda)F^(mu nu)F^(omega lambda)\epsilon_{\mu \nu \omega \lambda} F^{\mu \nu} F^{\omega \lambda} is also zero.
The result shows that for a plane wave solution of Maxwell's equations, the electric and magnetic fields oscillate in such a way that their magnitude is always equal (if multiplying the magnetic field with c^(2)c^{2} to get the correct units, the invariant F_(mu nu)F^(mu nu)F_{\mu \nu} F^{\mu \nu} is proportional to E^(2)-c^(2)B^(2)\boldsymbol{E}^{2}-c^{2} \boldsymbol{B}^{2} ) and that they are always orthogonal (epsilon_(mu nu omega lambda)F^(mu nu)F^(omega lambda):}\left(\epsilon_{\mu \nu \omega \lambda} F^{\mu \nu} F^{\omega \lambda}\right. is proportional to {:E*B)\left.\boldsymbol{E} \cdot \boldsymbol{B}\right).
1.122
a) Using the Lorentz invariant epsilon_(mu nu lambda omega)F^(mu nu)F^(lambda omega)=-8cE*B\epsilon_{\mu \nu \lambda \omega} F^{\mu \nu} F^{\lambda \omega}=-8 c \boldsymbol{E} \cdot \boldsymbol{B}, one finds
where epsilon^(mu nu lambda omega)\epsilon^{\mu \nu \lambda \omega} is a totally antisymmetric fourth rank tensor with epsilon^(0123)=1\epsilon^{0123}=1. For any vector index mu\mu, the inner product A^(mu)B_(mu)A^{\mu} B_{\mu} is Lorentz invariant, and for a Lorentz transformation Lambda\Lambda with det Lambda=1\operatorname{det} \Lambda=1, it holds that epsilon^('mu nu lambda omega)=epsilon^(mu nu lambda omega)\epsilon^{\prime \mu \nu \lambda \omega}=\epsilon^{\mu \nu \lambda \omega}. Thus, one has
{:(3.636)E^(')*B^(')=E*B:}\begin{equation*}
\boldsymbol{E}^{\prime} \cdot B^{\prime}=E \cdot B \tag{3.636}
\end{equation*}
b) The expression epsilon_(alpha beta gamma delta)F^(alpha beta)F^(gamma delta)\epsilon_{\alpha \beta \gamma \delta} F^{\alpha \beta} F^{\gamma \delta} is Lorentz invariant and equal to -8cE*B-8 c \boldsymbol{E} \cdot \boldsymbol{B}. Thus, if E*B=0\boldsymbol{E} \cdot \boldsymbol{B}=0 for one observer, then it is zero for any observer in inertial frames.
c) From Problem 1.120 we have E^(i)=i(k^(i)epsi^(0)-k^(0)epsi^(i))e^(ik*x)E^{i}=i\left(k^{i} \varepsilon^{0}-k^{0} \varepsilon^{i}\right) e^{i k \cdot x} and B^(i)=iepsilon^(ijk)k^(j)epsi^(k)e^(ik*x)B^{i}=i \epsilon^{i j k} k^{j} \varepsilon^{k} e^{i k \cdot x}. This gives immediately that E*B=E^(i)B^(i)=0\boldsymbol{E} \cdot \boldsymbol{B}=E^{i} B^{i}=0, since epsilon^(ijk)\epsilon^{i j k} is antisymmetric.
d) The components of E xx B\boldsymbol{E} \times \boldsymbol{B} can be written as (E xx B)^(i)=e^(2ik*x)(k^(i)M+epsi^(i)N)(\boldsymbol{E} \times \boldsymbol{B})^{i}=e^{2 i k \cdot x}\left(k^{i} M+\varepsilon^{i} N\right) with M=k^(0)epsi^(2)-k*epsiepsi^(0)M=k^{0} \varepsilon^{2}-\boldsymbol{k} \cdot \varepsilon \varepsilon^{0} and N=k^(2)epsi^(0)-k^(0)k*epsiN=\boldsymbol{k}^{2} \varepsilon^{0}-k^{0} \boldsymbol{k} \cdot \varepsilon. Choose k=(k^(0),k^(0),0,0)k=\left(k^{0}, k^{0}, 0,0\right) and epsi=(0,0,1,0)\varepsilon=(0,0,1,0) or epsi=(0,0,0,1)\varepsilon=(0,0,0,1). Then, it holds that E xx B=k^(0)(k^(0),0,0)e^(2ik*x)\boldsymbol{E} \times \boldsymbol{B}=k^{0}\left(k^{0}, 0,0\right) e^{2 i k \cdot x},
showing that only the 1-component of E xx B\boldsymbol{E} \times \boldsymbol{B} is nonvanishing, and thus, the product is proportional to k=(k^(0),0,0)\boldsymbol{k}=\left(k^{0}, 0,0\right).
1.123
Using the Lorentz force law, we have
{:(3.637)(dp)/(dt)=-e(u xx B):}\begin{equation*}
\frac{d p}{d t}=-e(\boldsymbol{u} \times \boldsymbol{B}) \tag{3.637}
\end{equation*}
where ee is the electron charge and p=m_(0)cx^(˙)=m_(0)u gamma\boldsymbol{p}=m_{0} c \dot{\boldsymbol{x}}=m_{0} \boldsymbol{u} \gamma with gamma-=1//sqrt(1-u^(2)//c^(2))\gamma \equiv 1 / \sqrt{1-u^{2} / c^{2}}. From the Lorentz force law, we deduce that
which implies that p^(2)=m_(0)^(2)u^(2)gamma^(2)=\boldsymbol{p}^{2}=m_{0}^{2} u^{2} \gamma^{2}= const. Since |u||\boldsymbol{u}| is constant and the magnetic field B=(0,0,B)\boldsymbol{B}=(0,0, B) is symmetric around the zz-axis, we make the ansatz
which is automatically consistent with the initial condition u(t=0)=(u,0,0)\boldsymbol{u}(t=0)=(u, 0,0). Now, the Lorentz force law for the components is
{:(3.640)(dp_(x))/(dt)=-eBu_(y)","quad(dp_(y))/(dt)=eBu_(x)","quad(dp_(z))/(dt)=0:}\begin{equation*}
\frac{d p_{x}}{d t}=-e B u_{y}, \quad \frac{d p_{y}}{d t}=e B u_{x}, \quad \frac{d p_{z}}{d t}=0 \tag{3.640}
\end{equation*}
which after inserting the ansatz yields
{:(3.641)(dp_(x))/(dt)=-eBu sin(alpha t)","quad(dp_(y))/(dt)=eBu cos(alpha t)","quad(dp_(z))/(dt)=0:}\begin{equation*}
\frac{d p_{x}}{d t}=-e B u \sin (\alpha t), \quad \frac{d p_{y}}{d t}=e B u \cos (\alpha t), \quad \frac{d p_{z}}{d t}=0 \tag{3.641}
\end{equation*}
where c_(1),c_(2)c_{1}, c_{2}, and c_(3)c_{3} are integration constants. Consistency with p=m_(0)u gamma\boldsymbol{p}=m_{0} \boldsymbol{u} \gamma means that c_(1)=c_(2)=c_(3)=0c_{1}=c_{2}=c_{3}=0 and alpha=eB//(m_(0)gamma)\alpha=e B /\left(m_{0} \gamma\right). Thus, integrating the expression for u\boldsymbol{u}, we obtain the trajectory of the electron as
{:(3.643)x(t)=(um_(0)gamma)/(eB)(sin((eBt)/(m_(0)gamma)),-cos((eBt)/(m_(0)gamma)),0)+x_(0):}\begin{equation*}
\boldsymbol{x}(t)=\frac{u m_{0} \gamma}{e B}\left(\sin \frac{e B t}{m_{0} \gamma},-\cos \frac{e B t}{m_{0} \gamma}, 0\right)+x_{0} \tag{3.643}
\end{equation*}
where x_(0)x_{0} is an integration constant. This trajectory is the equation for a circle perpendicular to the zz-axis with center at x_(0)\boldsymbol{x}_{0} and radius r=|x-x_(0)|=um_(0)gamma//(eB)r=\left|\boldsymbol{x}-\boldsymbol{x}_{0}\right|=u m_{0} \gamma /(e B). The time for one revolution is t_(0)=2pim_(0)gamma//(eB)t_{0}=2 \pi m_{0} \gamma /(e B).
1.124
The two quantities E*B\mathbf{E} \cdot \mathbf{B} and E^(2)-c^(2)B^(2)\mathbf{E}^{2}-c^{2} \mathbf{B}^{2} are Lorentz invariants, where E*B=0\mathbf{E} \cdot \mathbf{B}=0 and E^(2)-c^(2)B^(2)=0\mathbf{E}^{2}-c^{2} \mathbf{B}^{2}=0, since E=(cB,0,0)\mathbf{E}=(c B, 0,0) and B=(0,B,0)\mathbf{B}=(0, B, 0). Therefore, one has E^(')*B^(')=0\mathbf{E}^{\prime} \cdot \mathbf{B}^{\prime}=0 and E^('2)-c^(2)B^('2)=0\mathbf{E}^{\prime 2}-c^{2} \mathbf{B}^{\prime 2}=0. Inserting E^(')=(0,2cB,cB)\mathbf{E}^{\prime}=(0,2 c B, c B) and B^(')=(0,B_(y)^('),B_(z)^('))\mathbf{B}^{\prime}=\left(0, B_{y}^{\prime}, B_{z}^{\prime}\right), one obtains
Solving this system of equations, one finds that B_(y)^(')=+-BB_{y}^{\prime}= \pm B and B_(z)^(')=∓2BB_{z}^{\prime}=\mp 2 B. Thus, the answer is B^(')=+-(0,B,-2B)\mathbf{B}^{\prime}= \pm(0, B,-2 B).
1.125
The two quantities E^(2)-c^(2)B^(2)\mathbf{E}^{2}-c^{2} \mathbf{B}^{2} and E*B\mathbf{E} \cdot \mathbf{B} are Lorentz invariants, where E^(2)-c^(2)B^(2)=\mathbf{E}^{2}-c^{2} \mathbf{B}^{2}=-2alpha^(2)-2 \alpha^{2} and E*B=0\mathbf{E} \cdot \mathbf{B}=0, since E=(alpha,-alpha,0)\mathbf{E}=(\alpha,-\alpha, 0) and B=(0,0,2alpha//c)\mathbf{B}=(0,0,2 \alpha / c), where alpha!=0\alpha \neq 0. Therefore, one has E^('2)-c^(2)B^('2)=-2alpha^(2)\mathbf{E}^{\prime 2}-c^{2} \mathbf{B}^{\prime 2}=-2 \alpha^{2} and E^(')*B^(')=0\mathbf{E}^{\prime} \cdot \mathbf{B}^{\prime}=0. Inserting E^(')=(0,0,2alpha)\mathbf{E}^{\prime}=(0,0,2 \alpha) and B^(')=(B_(x)^('),alpha//c,B_(z)^('))\mathbf{B}^{\prime}=\left(B_{x}^{\prime}, \alpha / c, B_{z}^{\prime}\right), one obtains
Solving this system of equations, one finds that B_(x)^(')=+-(alphasqrt5)/(c)B_{x}^{\prime}= \pm \frac{\alpha \sqrt{5}}{c} and B_(z)^(')=0B_{z}^{\prime}=0. Thus, it follows that B^(')=(+-alphasqrt5//c,alpha//c,0)\mathbf{B}^{\prime}=( \pm \alpha \sqrt{5} / c, \alpha / c, 0).
1.126
The two quantities E^(2)-c^(2)B^(2)\mathbf{E}^{2}-c^{2} \mathbf{B}^{2} and E*B\mathbf{E} \cdot \mathbf{B} are Lorentz invariants, where E*B=0\mathbf{E} \cdot \mathbf{B}=0 and E^(2)-c^(2)B^(2)=-2beta^(2)\mathbf{E}^{2}-c^{2} \mathbf{B}^{2}=-2 \beta^{2}. Therefore, one has E^(')-c^(2)B^(')=-2beta^(2)\mathbf{E}^{\prime}-c^{2} \mathbf{B}^{\prime}=-2 \beta^{2} and E^(')*B^(')=0\mathbf{E}^{\prime} \cdot \mathbf{B}^{\prime}=0. Inserting E^(')=(2beta,0,0)\mathbf{E}^{\prime}=(2 \beta, 0,0) and B^(')=(B_(x)^('),B_(y)^('),beta//c)\mathbf{B}^{\prime}=\left(B_{x}^{\prime}, B_{y}^{\prime}, \beta / c\right), one obtains
Solving this system of equations, one finds that B_(x)^(')=0B_{x}^{\prime}=0 and B_(y)^(')=+-sqrt5beta//cB_{y}^{\prime}= \pm \sqrt{5} \beta / c. Thus, it follows that B^(')=(0,+-sqrt5beta//c,beta//c)\mathbf{B}^{\prime}=(0, \pm \sqrt{5} \beta / c, \beta / c).
1.127
Observer AA measures the electric and magnetic fields to be E=(alpha,0,0)\mathbf{E}=(\alpha, 0,0) and B=\mathbf{B}=(alpha//c,0,2alpha//c)(\alpha / c, 0,2 \alpha / c), respectively, where alpha!=0\alpha \neq 0. The two quantities E^(2)-c^(2)B^(2)\mathbf{E}^{2}-c^{2} \mathbf{B}^{2} and E*B\mathbf{E} \cdot \mathbf{B} are Lorentz invariants, where E^(2)-c^(2)B^(2)=-4alpha^(2)\mathbf{E}^{2}-c^{2} \mathbf{B}^{2}=-4 \alpha^{2} and E*B=alpha^(2)//c\mathbf{E} \cdot \mathbf{B}=\alpha^{2} / c. Therefore, one has E^('2)-c^(2)B^('2)=-4alpha^(2)\mathbf{E}^{\prime 2}-c^{2} \mathbf{B}^{\prime 2}=-4 \alpha^{2} and E^(')*B^(')=alpha^(2)//c\mathbf{E}^{\prime} \cdot \mathbf{B}^{\prime}=\alpha^{2} / c. Inserting E^(')=(E_(x)^('),alpha,0)\mathbf{E}^{\prime}=\left(E_{x}^{\prime}, \alpha, 0\right) and B^(')=(alpha//c,B_(y)^('),alpha//c)\mathbf{B}^{\prime}=\left(\alpha / c, B_{y}^{\prime}, \alpha / c\right), one obtains
Solving this system of equations, one finds that E_(x)^(')=-alphaE_{x}^{\prime}=-\alpha and B_(y)^(')=2alpha//cB_{y}^{\prime}=2 \alpha / c. Thus, it holds that E^(')=(-alpha,alpha,0)\mathbf{E}^{\prime}=(-\alpha, \alpha, 0) and B^(')=(alpha//c,2alpha//c,alpha//c)\mathbf{B}^{\prime}=(\alpha / c, 2 \alpha / c, \alpha / c), which are the electric and magnetic fields as measured by observer BB.
The electric and magnetic fields E^(')=(-alpha,alpha,0)\mathbf{E}^{\prime}=(-\alpha, \alpha, 0) and B^(')=(alpha//c,2alpha//c,alpha//c)\mathbf{B}^{\prime}=(\alpha / c, 2 \alpha / c, \alpha / c) imply that the electromagnetic field strength tensor as seen by observer BB is given by
The electromagnetic field strength tensor as seen by observer CC is then given by F^('')=LambdaF^(')Lambda^(T)F^{\prime \prime}=\Lambda F^{\prime} \Lambda^{T}, where
Here beta=beta(v)-=(v)/(c)\beta=\beta(v) \equiv \frac{v}{c} and gamma=gamma(v)-=(1)/(sqrt(1-beta^(2)))=(1)/(sqrt(1-(v^(2))/(c^(2))))\gamma=\gamma(v) \equiv \frac{1}{\sqrt{1-\beta^{2}}}=\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}. Thus, one obtains